\(\left(1+x^{2}\right)^{2}(1+x)^{n}\)
\(=\left(1+2 x^{2}+x^{4}\right)\left(1+n x+\frac{n(n-1)}{2} x^{2} \ldots \ldots\right)\)
\(\therefore a_{0}=1 ; a_{1}=n ; a_{2}=\frac{(n)(n-1)}{2}+2\)
\(a_{0}, a_{1}, a_{2} \text { are in A.P }\)
\(\therefore 1+\frac{(n)(n-1)}{2}+2 \quad=2 n\)
\(2+n^{2}-n+4=4 n\)
\(n^{2}-5 n+6 \quad=0\)
\((n-2)(n-3)=0\)
\(n =2 \text { (or) } 3 \)

\(\mathrm{AM} \geq \mathrm{GM} \)
\(\Rightarrow a \geq g \)

\({ }^{20} \mathrm{C}_{10}>{ }^{20} \mathrm{C}_{\mathrm{r}} \text { for all possible value of } \mathrm{r} \text {. }\)

\((2 x+3 y)^{20} \text { the term containing } x^{8} y^{12} \text { is }\)
\({ }^{20} C_{12}(2 x)^{20-12}(3 y)^{12}={ }^{20} C_{12} 2^{8} \cdot 3^{12} \cdot x^{8} y^{12}\)
\(\left(\because{ }^{20} \mathrm{C}_{12}={ }^{20} \mathrm{C}_{8}\right)\)
\(\text { Coefficient of } x^{8} y^{12} \text { is }{ }^{20} \mathrm{C}_{8} 2^{8} 3^{12}\)

\((2+2 x)^{10} \text { Term containing } x^{6} \text { is }\)
\({ }^{10} \mathrm{C}_{6}(2)^{10-6}(2 x)^{6}={ }^{10} \mathrm{C}_{6} 2^{4} 2^{6} x^{6}\)
\(\text { Coefficient of } x^{6} \text { is }{ }^{10} \mathrm{C}_{6} 2^{10}\)

\(e=1+\frac{1}{\lfloor 1}+\frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\ldots \ldots\)
\(e^{-1}=1-\frac{1}{\lfloor 1}+\frac{1}{\lfloor 2}-\frac{1}{\lfloor 3}+\frac{1}{4}-\ldots\)
\(\frac{e+e^{-1}}{2}=1+\frac{1}{\lfloor 2}+\frac{1}{\lfloor 4}+\ldots\)
\(\frac{1}{\lfloor 2}+\frac{1}{\lfloor 4}+\ldots . \quad=\frac{e+e^{-1}}{2}-1\)
\(=\frac{\mathrm{e}-2+\mathrm{e}^{-1}}{2} \)
\(=\frac{(\mathrm{e}-1)^{2}}{2 \mathrm{e}} \)
 

\(\mathrm{e}^{-2 x}=1-\frac{2 x}{1 !}+\frac{(2 x)^{2}}{2 !}-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{4}}{4 !}-\frac{(2 x)^{5}}{5 !}+\ldots\)
\(\text { Coefficient of } x^{5} \text { is } \frac{-2^{5}}{5 !}=\frac{-32}{120}=\frac{-4}{15}\)

\(a=6, S_{\infty}=18, r=? \)
\(S_{\infty} =\frac{a}{1-r}=\frac{6}{1-r}=18 \)
\(6 =18-18 r \)
\(18 r =12 \)
\(r =\frac{2}{3} \)

\(\mathrm{a} =1, \quad \mathrm{~d}=6, \quad \mathrm{r}=\frac{1}{2} \)
\(\mathrm{~S}_{\infty} =\frac{a}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1+\mathrm{r})^{2}} \)
\(=\frac{1}{1-\frac{1}{2}}+\frac{6 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^{2}} \)
\(=2+(3 \times 4)=14 \)

\(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32} \ldots . . =\sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2} . \)
\(=\sqrt{2}[1+2+3+\ldots .] \)
\(S_{n} =\frac{\sqrt{2}[n(n+1)]}{2} \)
\(=\frac{n(n+1)}{\sqrt{2}} \)

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

\(n^{\text {th }} \text { term is } \frac{n^{2}-n+2}{2}\)

\((38)^{5} =(39-1)^{15} \)
\(=(39)^{15}-{ }^{15} C_{1}(39)^{14}+\ldots+{ }^{15} C_{14}(39)-1 \)
The remainder will be 12 because all the terms except the last is divisible by 39 and so by 13, -1 remains

\(\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{n}}{2}[2 a+(\mathrm{n}-1) \mathrm{d}] \)
\(\mathrm{S}_{\mathrm{n}-1} =\frac{\mathrm{n}-1}{2}[2 a+(\mathrm{n}-2) \mathrm{d}] \)
\(\mathrm{S}_{\mathrm{n}-2} =\frac{\mathrm{n}-2}{2}[2 a+(\mathrm{n}-3) \mathrm{d}] \)
\(\mathrm{S}_{\mathrm{n}}-2 \mathrm{~S}_{\mathrm{n}-1}+\mathrm{S}_{\mathrm{n}-2} =2 a\left[\frac{\mathrm{n}-2(\mathrm{n}-1)}{2}+\frac{(\mathrm{n}-2)}{2}\right]+\frac{\mathrm{d}}{2}[\mathrm{n}(\mathrm{n}-1)-2(\mathrm{n}-1)(\mathrm{n}-2)+(\mathrm{n}-2)(\mathrm{n}-3)] \)
\(=2 a\left[\frac{\mathrm{n}-2 \mathrm{n}+2+\mathrm{n}-2}{2}\right]+\frac{\mathrm{d}}{2}\left[\mathrm{n}^{2}-\mathrm{n}-2 \mathrm{n}^{2}+6 \mathrm{n}-4+\mathrm{n}^{2}-5 \mathrm{n}+6\right] \)
\(= 0+\frac{\mathrm{d}}{2}(2)=\mathrm{d} \)

\(\mathrm{AM}=16, \quad \mathrm{GM}=8, \quad \mathrm{HM}=?\)
\(\frac{a+b}{2}=16 \Rightarrow a+b=32\)
\(\sqrt{a b}=8 \Rightarrow a b=64\)
\(\therefore \mathrm{HM} =\frac{2 a b}{a+b} \)
\(=\frac{2(64)}{32}= 4 \)

\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2 \sqrt{2}}, \ldots \ldots \text { form } a \text { HP. }\)

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)

Let a be the first term and r the common ratio of the G.P.
∴ S= a + ar + ar²+ ...+ ar^{n - 1}
\(=\frac{a(1-r^{n})}{1-r}\)  --- (1)
p =\(a\times ar \times ar^{2}\times...\times ar^{n-1}=a^{n}r^{1+2+3}+..+(n-1)=a^{n}r^{n{(n-1})/2}\)
∴ \(P^{2}=a^{2n}r^{n(n-1)}\)   --- (2)
\(R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^{2}}+....+\frac{1}{ar^{n-1}}\)
⇒ \(R=\frac{1}{a}.\frac{(1-\frac{1}{r^{n}})}{(1-\frac{1}{r})}=\frac{(r^{n}-1)}{(r-1)}.\frac{1}{ar^{n-1}}\)   [∵Here, r<1]
∴ \(\frac{S}{R}=a\frac{(1-r^{n})}{1-r}.\frac{r-1}{r^{n}-1} ar^{n-1}= a^{2}r ^{n-1}\)
∴ \((\frac{S}{R})^{n}=a^{2n}r^{n(n-1)}\)   --- (3)
From (2) and (3) we get \(P^{2}=(\frac{S}{R})^{n}\)

Let a be the first term and r be the common ratio of G.P.
∴ \(S_{1}=\frac{a(r^{n}-1)}{r-1}, S_{2}=\frac{a(r^{2n}-1)}{r-1}, S_{3}=\frac{a(r^{3n}-1)}{r-1}\)
where r ≠ 1
\(S_{3}-S_{2}= \frac{a}{r-1}(r^{3n}-r^{2n})=\frac{a(r^{n}-1)}{r-1}r^{2n}\)
\(S_{1}(S_{3}-S_{2})\frac{a(r^{n}-1)}{r-1}\times \frac{a(r^{n}-1)}{r-1}r^{2n}\)
=\([\frac{a(r^{n}-1)}{r-1}.r^{n}]^{2}\)   --- (1)
\((S_{2}-S_{1})=\frac{a}{r-1}(r^{2n}-r^{n})=\frac{a(r^{n}-1)}{r-1}r^{n}\)   --- (2)
∴ \(S_{1}(S_{3}-S_{2})=(S_{2}-S_{1})^{2}\) [From (1) and (2)]
When r = 1, S₁ = na, S₂ = 2na and S3 = 3 na
Then, \((S_{2}-S_{1})^{2}=2(na-na)^{2}=n^{2}a^{2}\) and \(S_{1}(S_{3}-S_{2})=na(3na-2na)\)
= na(na) = n² a²
∴ S₁ (S₃ - S₂) = (S₂ - S₁)² 

Let the consecutive coefficients ⁿC_r , ⁿC_r+1 and ⁿC_r+2 be the coefficients of T_r+1 T_r+2 and T_r+3 then
ⁿC_r+1:ⁿC_r+2:ⁿC_r+2 = 6:33:110.
Now,  \(\frac{^{n}C_{r}}{^{n}C_{r+1}} =\frac{6}{33}\)
⇒ \(\frac{n!}{r!(n-r)!}\times \frac{(r+1)!(n-r-1)!}{n!}=\frac{2}{11}\)
⇒ \(\frac{r+1}{n-r}=\frac{2}{11}\)
⇒ 11t +11 = 2n - 2r
⇒ 2n -13r = 11    --- (1)
Again, \(\frac{^nC_{r+1}}{^{n}C_{r+2}}=\frac{33}{110}\)
⇒ \(\frac{n!}{(r+1)!(n-r-1)!}\times \frac{(r+2)!(n-r-2)!}{n!}=\frac{3}{10}\)
⇒ \(\frac{(r+2)(r+1)!(n-r-2)!}{(r+1)!(n-r-1)(n-r-2)!}=\frac{3}{10}\)
⇒ \(\frac{r+2}{n-r-1}=\frac{3}{10}\)
⇒ 10(r + 2) = 3(n-r-1)
⇒ 3n - 13r = 23   --- (2)
Subtracting (2) from (1), we get n = 12

In the expansion of  (1+x)²ⁿ, Number of terms = 2n + 1, which is odd
There is only one middle term, the \(\frac{(2n+1)+1}{2}\) 1^th i.e. (n + 1)^th term
∴ T_n+1 is the only middle term T_n+1 = ²ⁿC_nxⁿ
Coefficient of T_n+1 = ²ⁿC_n  --- (1)
In the expansion of (1 + x)^2n-1,
Number of terms = 2n - 1 +1== 2n, which is even,
There are two middle terms, the \(\frac{2n}{2}\)^th i.e. T_n and T_n+1
T_n = ^2n-1C_n-1x^n-1 and T_n+1 = ^2n-1C_nxⁿ 
Coefficient of T_n = ^2n-1C_n-1 and coefficient of T_n+1 = ^2n-1C_n
Sum of the coefficients of the two middle terms in the expansion of
(1+x)^2n-1 = ^2n-1C_n + ^2n-1C_n-1 = ²ⁿC_n  --- (2)
RHS of(1) = RHS of (2), [∵ ⁿC_r+ⁿC_r-1 = ⁿ⁺¹C_r].

Suppose a² = x and \(\sqrt{a^{2}-1}=y\)
∴ \(a^{2}+\sqrt{a+1}=x+y \) and \(a^{2}-\sqrt{a^{2}-1}=x-y\).
Using binomial theorem,
\((x+y)^{4}={^4C_{0}x^{4}+{^4C_{2}x^{3}y}+{^4C_{2}x^{2}y^{2}}+{^4C_{3}xy^{3}+{^4C_{4}y^{4}}}}\)
⇒ (x+y)⁴ = x⁴+ 4x³y + 6x²y² + 4xy³ + y⁴ and
(x-y)⁴ = ⁴C₀x⁴ - ⁴C₁x³y + ⁴C₂x²y² - ⁴C₃xy³ + ⁴C₄y⁴
⇒ \((x-y)^{4}=x^{4}-4x^{3}y+6x^{2}y^{2}-4xy^{3}+y^{4}\)
∴ \((x+y)^{4}+(x-y)^{4}=2[x^{4}+6x^{2}y^{2}+y^{4}]\)
\(\therefore (a^{2}+\sqrt{a^{2}+1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}=2[(a^{2})^{4}+6(a^{2})^{2}(\sqrt{a^{2}-1})^{2}+(\sqrt{a^{2}-1})^{4}] \)
= \(2[a^{8}+6a^{4}(a^{2}-1)+(a^{2}-1)^{2}]\)
= \(2[a^{8}+6a^{6}-6a^{4}+a^{4}-2a^{2}+1]\)
= \(2[a^{8}+6a^{6}-5a^{4}-2a^{2}+1]\)

We have, C (n, r - 2), C(n, r - 1), C(n, r) = 1 : 3 : 5
⇒ \(\frac{C(n,r-2)}{1}=\frac{C(n,r-1)}{3}=\frac{C(n,r)}{5}\)    --- (1)
From the first two members of (1), 3C(n, r - 2), = C(n, r - 1)
⇒ \(3 \frac{(n)!}{(r-2)!(n-r+2!)}=\frac{n!}{(r-1)!(n-r+1)!}\)
⇒ \(\frac{3}{(r-2)!(n-r+2)(n-r+1)!}=\frac{1}{(r-1)(r-2)!(n-r+1)!}\)
⇒ \(\frac{3}{n-r+2}=\frac{1}{r-1}\)
⇒ 3(r-1) = n - r + 2
⇒ n-4r + 5 = 0    --- (2)
From the last two members of (1),
5C (n, r, 1) = 3 C (n,r)
⇒ \(\frac{n!}{(r-1)!(n-r+1)!}=3\frac{n!}{r!(n-r)!}\)
⇒ \(\frac{5}{(r-1)!(n-r+1)(n-r)!}=\frac{3}{r(r-1)!(n-r)!}\)
⇒ \(\frac{5}{n-r+1}=\frac{3}{r} \Rightarrow 5r=3(n-r+1)\)
⇒ 3n - 8r + 3 = 0    --- (3)
Solving (2) and (3), we get n = 7 and r = 3.

\(\sqrt [ 3 ]{ x^3+7 } ={(x^3+7)}^{{1\over 3}}\)
\(={\left[ x^3\left( 1+{7\over x^3} \right) \right]}^{{1\over 3}}\)   (\(\left |{7\over x^3}\right |<1\) as x is large)
\(=x{\left( 1+{7\over x} \right)}^{1\over 3}\)
\(=x\left( 1+{1\over 3} \times {7\over x^3}+{{{1\over3}\left( {1\over 3}-1 \right)}\over{2!}} {\left( {{7\over x^3}} \right)}^{2} +......\right)\)
\(=x\left( 1+{7\over 3}\times{1\over x^3}-{49\over 9}\times{1\over x^6}+...... \right)\)
\(=x+{7\over 3}\times{1\over x^2}-{49\over 9}\times{1\over x^5}+...\)
\(\sqrt [ 3 ]{ x^3+4 } ={(x^3+4)}^{1\over 3}\)
\(={\left[ x^3\left( 1+{4\over x^3} \right) \right]}^{1\over 3}\)
\(=x\left( 1+{4\over x^3} \right)^{1\over 3}\)
\(=x{\left( 1+{1\over3}\times{4\over x^3}+{{1\over 3}\left( {1\over3}-1 \right)\over{2!}} {\left( {4\over x^3} \right)}^{2}+... \right)}^{1\over3}\)
\(=x+{4\over 3}\times{1\over x^3}-{16\over 9}\times{1\over x^5}+...\)
Since x is large, \({1 \over x}\) is very small and hence higher powers of \({1 \over x}\) are negligible.
Thus \(\sqrt [ 3 ]{ x^3+7 } =x+{7\over 3}\times{1\over x^2}\) and \(\sqrt [ 3 ]{ x^3+4 } =x+{4\over3}\times{1\over x^3}.\) Therefore
\(\sqrt [ 3 ]{ x^3+7 } -\sqrt [ 3 ]{x^3+4 } =\left(x+{7\over 3}\times{1\over x^2} \right)-\left( x+{4\over 3}\times{1\over x^2} \right)={1\over x^2}\)

We know that for |x| < 1
\((1+x)^n=1+nx+{n(n-1)\over 2!}x^2+{n(n-1)(n-2)\over3!}x^3+....\)
\(\sqrt [ 3 ]{ 65} ={65}^{{1\over3}}\)
\(={(64+1)}^{1\over 3}\)
\(={64}^{1\over 3}{\left( 1+{1\over 64} \right)}^{1\over3}\Rightarrow4{\left( 1+{1\over64} \right)}^{1\over 3}\)
\(=4\left( 1+{1\over 3}\times{1\over 64}+{{1\over 3}\left({1\over3 }-1\right)\over{2!}} \times {\left({1\over 64} \right)}^{2}+....... \right)\)
\(=4+{1\over 48}-4\times{1\over 9}\times{1\over 64}\times{1\over 64}+....\)
\(=4+{1\over 48}-{4\over 36864}+...\)
\(=4+{1\over48}-{1\over 9216}+....\)
\(\approx 4 + 0.02 \)      ( since \({1\over 9216}+...\) is very small )
\(\sqrt [ 3 ]{ 65 }=4.02\) (approximately)

(Clearly, we have to use the expansion of (1 +x)^-2. So, we have to write (3 + 2x) as \(3(1+{2x\over3})\) and proceed.)
\({1\over (3+2x)^2}={1\over 3^2(1+{2x\over3})^2}\)
\(={1\over 9}(1+{2x\over3})^{-2}\)
\(={1\over 9}(1+y)^{-2}\)    \((where \ y={2x\over 3})\)
\(={1\over 9}(1+2y+3y^3-4y^3+5y^4-....),\)   if |y|<1
\(={1\over 9}(1-2({2x\over3})+3({2x\over 3})^2-4({2x\over3})^3+5({2x\over3})^4-...),|{2x\over3}|<1\)
\(={1\over9}(1-{4\over3}x+{4\over3}x^2-{32\over27}x^3+{80\over81}x^4-......)\)
Thus,\({1\over (3+2x)^2}={1\over9}({4\over27}x+{4\over27}x^2-{32\over243}x^3+{80\over729}x^4-....,|x|<{3\over2})\)
The expansion is valid if |y| < 1. So, the expansion is valid if |x|<\(3\over2\)

S₁₀ = 52, S₁₅ = 77, S₂₀ =?
\({S}_{10}={10\over 2}(2a(10-1)d)=52\)
= 5 (2a + 9d) = 52
\(2a+9d={52\over 5}\)
\({S}_{15}={15\over 2}(a+(15-1)d)=77\)
\(={15\over 2}(2a+14d)=77\)
\(2a+14d={154\over 15}\)
\(2a+9d={52\over 2}\)
(2)-(1) \(\Rightarrow\)    \({2a+14d={154\over 15}\over 5d={154\over 15}-{52\over 2}={154-156\over 15}}\)
\(5d={-2\over 15}\Rightarrow d={-2 \over 75}\)
substituting in (1) ​​​​​​​\(\Rightarrow\) \(2a-{18\over 75}={52\over 5}\)
\(2a={52\over 5}+{18\over 75}={780+18\over 75}\)
\(2a={798\over 75}\Rightarrow a={399\over 75}={133 \over 25}\)
\({S}_{20}={20 \over 2}\left[ \left( {798 \over 75} \right)+19\left( {-2\over 75} \right) \right]\)
\(=10{(798-38)\over 75}=10\left( {760\over 75} \right)\)
\(={1520\over 15}={304\over 3}\)

Let t_k denote the k^th term of the given series.
Then \(t_k{1\over k(k+1)}.\) 
By using partial fraction we get
\({1 \over k(k+1)}={1\over k}-{1\over{k+1}}\)
Thus \(t_1+t_2+....+t_n=\left( 1-{1\over 2} \right)+\left( {1\over 2}+{1\over 3} \right)+\left( {1\over 3}-{1\over 4} \right)+...+\left( {1\over n} -{1\over n+1} \right)=1-{1\over n+1}.\)

Let t_k denote the k^th term of the given series.
Then \(t_k={1\over \sqrt{k}+\sqrt{k+1}}.\) 
If we are successful in writing the k^th term as a difference of two expressions, then we can solve using this technique.
We have
\(t_k={1\over\sqrt{k}+\sqrt{k+1}}={{\sqrt{k}-\sqrt{k+1}}\over{(\sqrt{k}+\sqrt{k+1})(\sqrt{k}-)\sqrt{k+1}}}={{\sqrt{k}-\sqrt{k+1}}\over{k-(k+1)}}=\sqrt{k+1}-\sqrt{k}\)
Thus, \(t_1+t_2+....+t_4=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{n-1}-\sqrt{n})=\sqrt{n+1}-1\)

Rs.270

\(\frac{x}{8}\)

\(\frac{x}{4}\)

\(\frac{2^n+3^n}{n}\)

Let a_n denote the n^thterm of the given series.
Then a_n \(={1\over n^2+5n+6}\) By using partial fraction, we get a_{n \(={1\over n+2}-{1\over n+3}\)}
Let S_n denote the sum of first n terms of the given series. Then
\(S_n=a_1+a_2+...+a_n=({1\over3}-{1\over 4})+({1\over4}-{1\over 5})+({1\over5}-{1\over 6})+....+({1\over n+2}-{1\over n+3})={1\over 3} -{1\over n+3}\)
But as n tends to infinity, \({1\over n+3}\) tends to zero and hence \({1\over 3}-{1\over n+3}\) tends to \({1\over 3}\). In other words S_n tends to \({1\over 3}\)
Thus\(\sum _{ n=1 }^{\infty }{1\over n^2+5n+6 } ={1\over 3}\)

We have 7⁴⁰⁰ = (7²)²⁰⁰ = (50 - 1)²⁰⁰
= ²⁰⁰C₀ 50²⁰⁰ - ²⁰⁰C₁ 50¹⁹⁹ + .... + ²⁰⁰C₁₉₈ 50²(-1)¹⁹⁸ + ²⁰⁰C₁₉₉ 50(-1)¹⁹⁹ + ²⁰⁰C₂₀₀(-1)²⁰⁰
= 50²-(²⁰⁰C₀50¹⁹⁸ - ²⁰⁰C₁50¹⁹⁷ + ..... + ²⁰⁰C₁₉₈(-1)¹⁹⁸) - 200 \(\times\) 50 + 1
As 50² and 200 are divisibie By 100, the last two digits: 0, 1.

\((x^2+\sqrt{1-x^2})^5\) = ⁵C₀(x²)⁵ \({(\sqrt{1-x^2})}^{0}\) + ⁵C₁(x²)⁴\({(\sqrt{1-x^2})}^{1}\)+⁵C₂(x²)³\({(\sqrt{1-x^2})}^{2}\)+⁵C₃(x²)²\({(\sqrt{1-x^2})}^{3}\)+⁵C₄(x²)¹\({(\sqrt{1-x^2})}^{4}\)+⁵C₅(x²)⁰\({(\sqrt{1-x^2})}^{5}\)
= x¹⁰+5x²\(\sqrt{1-x^2}\) +10x⁶(1-x²)+10x⁴(1-x²)\(\sqrt{1-x^2}+5x^2(1-x^2)^2+(1-x^2)^2(\sqrt{1-x^2})\)
\((x^2-\sqrt{1-x^2})^5\) =⁵C₀(x²)⁵\({(\sqrt{1-x^2})}^{0}\) - ⁵C₁(x²)⁴ \((\sqrt{1-x^2})^1\) +⁵C₂(x²)³\({(\sqrt{1-x^2})}^{2}\) -⁵C₃(x²)²\({(\sqrt{1-x^2})}^{3}\)+⁵C₄(x²)¹\({(\sqrt{1-x^2})}^{4}\) -⁵C₅(x²)⁰\({(\sqrt{1-x^2})}^{5}\)
\(={x}^{10}-5x^8\sqrt{1-x^2}+10x^6(1-x^2)-10x^4(1-x^2)\sqrt{1-x^2}+5x^2(1-x^2)^2-(1-x^2)^2(\sqrt{1-x^2})\)
Thus 
\((x^2+\sqrt{1-x^2})^5+(x^2+\sqrt{1-x^2})^5\) = 2[x¹⁰ + 10x⁶ (1 - x²) + 5x² (1-x²)²]
= 2[x¹⁰ + 10x⁶ -10x⁸+ 5x² (1- 2x² +x⁴)]
= 2[x¹⁰ - 10x⁸ + 15x⁶ - 10x⁴ + 5x²]

To prove this it is enough to prove, 6ⁿ - 5n = 25k + 1 for some integer k. We first consider the expansion
(1+ x )ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² + ... + ⁿC_n-1 x^n-1 + ⁿC_n xⁿ, \(n \in N\)
Taking x = 5 we get (1 + 5)ⁿ = ⁿC₀ + ⁿC₁ 5 + ⁿC₂ 5² + ... + ⁿC_n-1 5^n-1 + ⁿC_n5n.
The above equality reduces to 6ⁿ = 1 + 5n + 25 (ⁿC₂ + 5ⁿC₃ + ... + ⁿC_n 5^n-2).
That is, 6ⁿ - 5n = 1 + 25(ⁿC₂ + 5 ⁿC₃ + ... + ⁿC_n 5^n-2) = 1 + 25k, \(k\in N\)
Thus 6ⁿ - 5n always leaves remainder 1 when divided by 25 for all positive integer n.

It is given that T₂ = 240, T₃ =·720 and T₄ = 1080.
T₂ = ⁿC₁ x^n-1 = 240 ...(1)
T₃ = ⁿC₂ x^n-2a² = 720 ....(2)
T₄ = ⁿC₃ x^n-3 a³ = 1080 ....(3)
Dividing (2) by (1) and (3) by (2) we get
\({a\over x}={6\over n-1}\)  ...(4)
\({a \over x}={9\over2(n-2)}\) .....(5)
From (4) and (5)    \({6\over n-1}={9\over 2(n-2)}\)
Thus n = 5. Substituting n = 5 in (1), (4) and dividing (1) by (4)
\({{5x^4 a}\over{{{a}\over{x}}}}={{240}\over{{6}\over{4}}}\)
Thus 5x⁵ = 160 and hence x = 2. Substituting in (4) we get a = 3.

Let the numbers be a and b
\(∴ A={a+b\over 2}, G=\sqrt ab\ and\ H={2ab\over a+b}\)
Given A-G = 10 and A - H = 16
G = A-10 and H = A-16
We know G² = AH
⇒ (A - 10)² = A (A-16)
⇒ A² + 100 - 20A = A²-16A
⇒ 100 = \(4A⇒A=25⇒{a+b\over2}=25⇒a+b=50\)
∴ G = A-10 = 25 - 10 = 15
\(∴ \sqrt{ab}=15⇒ab=225\)
\(⇒ b={225\over a}\)

Substituting (2) in (1) we get,
\(a+{225\over a}=50\)
\(⇒\ {a^2+225\over a}=50\)
⇒ a² + 225 = 50a
⇒ a² - 50a + 225 = 0
⇒ (a - 45) (a -5) = 0
⇒ a = 5, 45
If a = 5, b = \({225\over 5}=45\)
If a = 45, b = \({225\over 45}=45\)
Hence the numbers are 5, 45

a = k^x , b = k^y and c = k^z
Also, given that a,b, c are in GP
\(\frac { b }{ a } =\frac { c }{ b } \Rightarrow { b }^{ 2 }=ac\)
ky² = k^x .k^z
k²y = k^x+z
2y  = x +z
y + y = x + z
y - x = z - y
Common difference is same for x, y, z
∴ x, y, z are in arithmetic progression

Since \(\alpha ,\beta \) are the roots of the equation \(​​{ x }^{ 2 }-px+q=0\),we have
\(\alpha +\beta =-\frac { \left( -p \right) }{ 1 } =p\quad and\quad \alpha \beta =\frac { q }{ 1 } q\)
\(\therefore \log { (1+px+q{ x }^{ 2 }) } =\log { \left[ 1+(\alpha +\beta )x+\alpha \beta { x }^{ 2 } \right] } \)
\(=\log { \left[ (1+\alpha x)(1+\beta x) \right] } \)
\(=\log { (1+\alpha x)+log(1+\beta x) } \)
\(=\left( \alpha x-\frac { { \alpha }^{ 2 } }{ 2 } { x }^{ 2 }+\frac { { \alpha }^{ 3 } }{ 2 } { x }^{ 3 }+......+\infty \right) +\left( \beta x-\frac { { \beta }^{ 2 }{ x }^{ 2 } }{ 2 } +\frac { { \beta }^{ 3 }{ x }^{ 3 } }{ 3 } +.....\infty \right) \)
\(=(\alpha +\beta )x-\frac { { (\alpha }^{ 2 }+\beta ^{ 2 }) }{ 2 } { x }^{ 2 }+\frac { { (\alpha }^{ 3 }+\beta ^{ 3 }) }{ 3 } { x }^{ 3 }-...\infty \)
Hence proved. 

Given t_k  is the K^th term of G.P 
\( t_{n} =a r^{n-1} \)
\(t_{n-k} =a r^{n-k-1} \)
\(t_{n+k} =a r^{n+k-1} \)
\(\therefore r_{1} =\frac{t_{2}}{t_{1}} \)
\(r_{1} =\frac{t_{n}}{t_{n-k}}=\frac{a r^{n-1}}{a r^{n-k-1}} \)
\(r_{1} =\frac{t_{n}}{t_{n-k}}=r^{k}\)....................(1)
\( r_{2} =\frac{t_{3}}{t_{2}} \)
\(\Rightarrow r_{2} =\frac{t_{n+k}}{t_{n}}=\frac{a r^{n+k-1}}{a r^{n-1}} \)
\( =r^{n+k-1-n+1}\)...............(2)
From (1) and (2) r₁ = r₂
Since the common ratio is same, t_n-k, t_n, t_n+k

Given \({a\over r}.a.ar=5832\)
⇒ a³ = 5832 =  18³
⇒ a =18
Also given \({a\over r},a+6, ar+9\) form an A.P
\(∴ a+6-{a\over r}=ar+9-a-6\)
\(⇒\ 18 + 6 --{18\over r}=18r + 9 - 18 - 6\ \ \ [∵ a = 18 ]\)
\(⇒\ 24-{18\over r}=18r-15\)
\(⇒ 24 + 15=18r+{18\over r}\)
\(⇒ 39 ={18r^2+18\over r}\)

⇒ 39r =  18r² + 18
⇒ (2r- 3) (3r- 2) =0
⇒ \(r={3\over2},{2\over 3}\)
Case (i) When \(a=18,r={3\over 2},\)the numbers m G.P. are \({18\over {3\over 2}},18,18\left(2\over 3\right)⇒27,18,12\)
Case (ii) When \(a=18,r={2\over 3},\) the numbers m G.P. are \({18\over {2\over 3}},18,18\left(2\over 3\right)⇒27,18,12\)

\(\frac { 1-4x-{ x }^{ 2 } }{ ex } \) = (1-4x-x²)e^-x
(1-4x-x²) =\(\left( 1+\frac { (-x) }{ 1! } +\frac { { \left( -x \right) }^{ 2 } }{ 2! } +\frac { { \left( -x \right) }^{ 3 } }{ 3! } +.....\infty \right) \)
(1-4x-x²) =\(\left( 1-x\frac { { \left( x \right) }^{ 2 } }{ 1! } +\frac { { \left( x \right) }^{ 3 } }{ 2! } +\frac { { \left( x \right) }^{ 4 } }{ 3! } +\frac { { \left( x \right) }^{ 5 } }{ 4! } +.....\infty \right) \)
Getting the x⁵ term alone,
\(=\frac { { -x }^{ 5 } }{ 5! } -4x\times \frac { { x }^{ 4 } }{ 4! } +{ x }^{ 2 }\times \frac { { x }^{ 3 } }{ 3! } =\left( -\frac { 1 }{ 120 } -\frac { 4 }{ 24 } +\frac { 1 }{ 6 } \right) { x }^{ 5 }=-\frac { 1 }{ 120 } { x }^{ 5 }\)
\(\therefore\) co-efficient of  x⁵ is \(\frac {- 1 }{ 120 } \)

Fourth root of 623 = \(\left( \frac { -2 }{ 625 } \right) \)
\(\left( \frac { -2 }{ 625 } \right) \)
\(={ \left[ 625\left( 1-\frac { 2 }{ 625 } \right) \right] }^{ \frac { 1 }{ 4 } }=5{ \left[ 1+\left( -\frac { 2 }{ 625 } \right) \right] }^{ \frac { 1 }{ 4 } }\)
\(={ 5\left[ 1+\frac { 1 }{ 4 } \left( \frac { -2 }{ 625 } \right) +\frac { \frac { 1 }{ 4 } \left( -\frac { 3 }{ 4 } \right) }{ 1.2 } { \left( \frac { -2 }{ 625 } \right) }^{ 2 } \right] }\)
Other terms will have more than seven zeroes after the decimal]
= 5[1-0.0008 - 0.0000009]
\(\sqrt [ 4 ]{ 623 } =4.9959955\)

Let T_n be the nth term of the given series.
\(\therefore { T }_{ n }=\frac { { 1 }^{ 3 }+{ 2 }^{ 3 }+3^{ 3 }+....+n\quad terms }{ 1+3+5....+n\quad terms } =\frac { \Sigma { n }^{ 3 } }{ \frac { n }{ 2 } (2(1)+(n-1)2) } \)
\(\frac { { \left( \frac { n(n+1) }{ 2 } \right) }^{ 2 } }{ \frac { n }{ 2 } (2n) } \left[ \because \Sigma { n }^{ 3 }={ \left( \frac { n(n+1) }{ 2 } \right) }^{ 2 } \right] \)
\(=\frac { { n(n+1) }^{ 2 } }{ \frac { n }{ 2 } (2n) } =\frac { 1 }{ 4 } ({ n }^{ 2 }+2n+1)\)
\(=\frac { { n(n+1) }^{ 2 } }{ \frac { n }{ 2 } (2n) } =\frac { 1 }{ 4 } ({ n }^{ 2 }+2n+1)\)
\(=\frac { 1 }{ 4 } (\Sigma { n }^{ 2 }+\Sigma 2n+n)=\frac { 1 }{ 4 } \left[ \frac { n(n+1)(2n+1) }{ 6 } +\frac { 2.n(n+1) }{ 2 } +n \right] \)
\(=\frac { 1 }{ 4 } \left[ \frac { n(n+1)(2n+1)+6n(n+1)+6n }{ 6 } \right] \)
\(=\frac { n }{ 24 } \left[ { 2n }^{ 2 }+9n+13 \right] =\frac { n({ 2n }^{ 2 }+9n+13) }{ 24 } \)

Given series is \(\sum_{n=1}^{\infty}{1\over 2^{n-1}}\left({1\over 9^{n-1}}+{1\over9^{2n-1}}\right)\)
\(S_\infty=1\left(1+{1\over9}\right)+{1\over2}\left( {1\over9}+{{1\over9^3}} \right)+{1\over 2^2}\left( {1\over9^2}+{1\over9^5} \right)+...\)
\(=\left( 1+{1\over9} \right)+{1\over2}\left( {1\over9}+{1\over 9^3} \right)+{1\over2^2}\left( {1\over2^2}+{1\over 9^5} \right)+..\)
Separating the first term and second term from each bracket, we get
\(=\left[ 1+{1\over 2}+{1\over 9}+{1\over 2^2}\left( {1\over 9^2}\right)+..\right]+\left[ {1\over9}+{1\over2}\left( {1\over 9^5} \right)+...\right]\)
\(=\left[1+{1\over2}\left(1\over 9\right)+{1\over 2^2}\left( {1\over 9^2} \right)+...\right]+{1\over9}\left[1+{1\over2}\left( {1\over 9^2}\right) +{1\over2^2} \left( {1\over 9^4} \right)+...\right]\)
Now consider \({1+{1\over2}}\left(1\over9\right)+{1\over 2^2}\left(1\over 9^2\right)+...\)
Here a = 1, \(r={1\over 18}\)
\(∴\ S_\infty={a\over 1-r}={1\over 1-{1\over 8}}={18\over 17}\)
In \(1+{1\over2}\left(1\over9^2\right)+{1\over 2^2}\left(1\over 9^4\right)+...\)
\(a=1,r={1\over 81\times2}={1\over162}\)
\(∴\ S_\infty =1, r={a\over 1-r}={1\over 1-{1\over 162}}={162\over 161}\)
Substituting these values in (1) we get
\(S_\infty={18\over 17}+{1\over 9}\times{162\over161}={18\over17}+{162\over 1449}={26082+2754\over 17\times1499}={28836\over 24633}\)
\({ S }_{ \infty }=1.170\)

\(\frac { 3-4x+{ x }^{ 2 } }{ { e }^{ 2x } } =\left( 3-4x+{ x }_{ 2 } \right) { e }^{ -2x }\)
\(=\left( 3-4x+{ x }^{ 2 } \right) \left( 1-\frac { 2x }{ 1! } +\frac { { \left( 2x \right) }^{ 2 } }{ 2! } -\frac { { \left( 2x \right) }^{ 3 } }{ 3! } +\frac { { \left( 2x \right) }^{ 4 } }{ 4! } -... \right) \)
\(=\left( 3-4x+{ x }^{ 2 } \right) \left( 1-2x+\frac { { 4x }^{ 2 } }{ 2 } -\frac { { 8 }^{ 3 } }{ 6 } +\frac { { 16 }^{ 4 } }{ 24 } -... \right) \)
\(=\left( 3-4x+{ x }^{ 2 } \right) \left( 1-2x+{ 2x }^{ 3 }-\frac { { 4x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 4 } }{ 3 } -... \right) \)
\(=2+\frac { 16 }{ 3 } +2=\frac { 6+16+6 }{ 3 } =\frac { 28 }{ 3 } \)
Hence coefficient of x⁴ is \(\frac { 28 }{ 3 } \)

The given series is 1 + 5(-1)+9(1)+13(-1)+....
= 1 + 5(-1)+9(-1)²+13(-1)³+....
This is an arithmetic - geometric series with corresponding A.P. 1, 5, 9, 13.... and G.P. 1, -1, (-1)², (-1)³, ....
\(\therefore\) Tn of A.G. series = (Tn of A.P) (Tn of G.P)
= [ 1 + ( n - 1)4 ]  [ 1 (-1)^n-1]  [ \(\because\) T_n in A.P. is a + (n - 1)d, T_n in G.P. is a r^n-1 ]
= (4n-3)(-1)^n-1
Let Sn be the Sum of the first n terms of AG series
S_n = T₁ + T₂ + T₃ + ... T_n-1 + Tn
S_n = 1 + 5(-1)+9(-1)² + .. + (4n-7)(-1)^n-2 + (4n-3) (-1)^n-1  ...(2)
Multiplying by (-1) we get,
-1 S_n = 1 + 5(-1)+9(-3)³ + .. + (4n-7)(-1)^n-1 + 4n-3(-1)ⁿ
(2)-(3) we get,
S_n + S_n = 1 + [4(-1) + 4(-1)²]+...+4(-1)^n-1]-(4n-3)(-1)ⁿ
\(\Rightarrow\)   \(2S_n=1+{4(-1)[1-(-1)^{n-1}]\over1-(-1)}-(4n-3)(-1)^n\)    \(\begin{bmatrix} \because For\ GP, S_N={a(1-r^n)\over 1-r} \\a=1,r=-1 \end{bmatrix}\)
\(\Rightarrow\)    \(2S_n=1-2()1-(-1)^{n-1}-(4n-3){(-1)}^{n}\)
\(\Rightarrow\)    \(S_n={1\over 2}-1+{(-1)}^{n-1}-{4n-3\over 2}{(-1)}^{n-1}(-1)\)
\(\Rightarrow\)   \(S_n={-{1\over 2}}+{2+4n-3\over 2}{(-1)}^{n-1}\)
\(\Rightarrow\)   \(S_n={1\over 2}[-1+(4n-14){(--1)}^{n-1}].\)

Given series is \(1+2\left( {1 \over 5} \right)+3{\left( {1\over 5} \right)}^{2}+4{\left({1\over 5} \right)}^{2}+....\)   ...(1)
This is an arithmetico - geometric series with corresponding A.P. 1, 2, 3, 4, .... and G.P: \(1,{1 \over 5},\left( {1\over 5} \right)^2,{\left( {1\over 5} \right)}^{3},......\)
Let \(S=1+2\left({1\over 5} \right)+3{\left( {1\over 5} \right)}^{2}+4{\left({1\over 5} \right)}^{3}+....\)  ...(2)
Multiplying both sides by the common ratio \(={1\over 5}\) of G.P. we get,
\({1\over 5}S={1\over5}+2{\left( {1 \over 5} \right)}^{2}+3{\left( {1\over 5} \right)}^{3}+4{\left( {1\over 5} \right)}^{4}+....\)
(2) - (1), we get,
\(S-{1\over5}S=1+(2-1)\left( {1\over5} \right)(3-2){\left( {1\over 5} \right)}^{2}+(4-3){\left({1\over 5} \right)}^{2}+...\)
\(\Rightarrow\)     \(S-{1\over 5}S=1+\left[ {1\over5}+{\left( {1\over 5} \right)}^{2}+{\left( {1\over 5} \right)}^{3}+... \right]\)
\(=1+\frac { \frac { 1 }{5 } }{ 1-\frac { 1 }{ 5 } } \)    \(\left[ \because {S}_{\infty}={a \over 1-r} \right]\)
\(=1+\frac { \frac { 1 }{ 5 } }{ \frac { 4 }{ 5 } } =1+{1\over 4}={5\over 4}\)   \(\left[ \because |{1\over 5}| ={1\over 5}<1\right]\)
\(\Rightarrow\)   \(S={5\over 4}\times{5\over 5}={25\over16}\)

Let the numbers be a and b.
\(\therefore\)  H.M \(={2ab \over a+b}\) and G.M \(=\sqrt{ab}\)
Given \({2ab\over a+b}:\sqrt{ab}=12:13\)
\(\Rightarrow\)    \({2ab \over (a+b)\sqrt{ab}}={ 12 \over 13}\)
\(\Rightarrow\)   \({\sqrt{ab}\over a+b }={12\over 13}\)
\(\Rightarrow\)    \({\sqrt{ab} \over a+b}={6\over 13}\Rightarrow6a+6b=13\sqrt{ab}\)
Dividing by b we get,
\(6\left( {a \over b} \right)+6=13\sqrt{a \over b}.\)
\(\Rightarrow\)   \(6\left( { a \over b} \right)-13\sqrt{{a \over b}}+6=0\)
\(\Rightarrow\)  \(\sqrt{a\over b}=13\pm\sqrt{169-144\over 12}\)    \(\begin{bmatrix}x = {-b \pm \sqrt{b^2-4ac} \over 2a} \\a=6,b=-13,c=6 \end{bmatrix}\)
\(\Rightarrow\ \ \sqrt{a\over b}={13+5\over 5}={2\over 3}\) or \({3\over 2}\)
\(\therefore\quad {a\over b}={4\over 9}\) or \({9\over 4}\)
\(\therefore\)   a : b = 4 : 9 or 9 : 4.

Let the H.P. be \({{1}\over{a}},{1\over a+d},{1\over a+2d},...\)
\(\therefore\)     \({T}_{1}={1\over a}={2 \over 5}\) and \({T}_{2}={ 1 \over a+d}={12 \over 13}\)
\(\Rightarrow\)  \(a={5\over 2}\) and \(a+d={13 \over 12}\)
\(\Rightarrow\)  \(a={5\over 2}\) and \({5\over 2}+d={13 \over 12}\Rightarrow d{13 \over 12}-{5\over 2}\Rightarrow d={13-20\over12 }={-17\over 12}\)
Now, \({T}_{n}-{1 \over {a+(n-1)d}}={{1}\over{{{5}\over{2}}+(n-1)\left( {{-17}\over{12}} \right)}}={12\over30-17n+17}\)
\(={12\over47-17n}\)
Tn is largest when 47 - 17n is least and positive
For n = 2, 47 - 17n is least and positive
\(\therefore\) Tn is largest when n = 2
\(\therefore\) The largest term \(={12\over 47-17(2)}={12\over13}.\)

Let the positive numbers be a, b. Let a > b.
\(\therefore A=\frac { a+b }{ 2 } \Rightarrow a+b=2A \ \text {and} G=\sqrt { ab } \Rightarrow ab\Rightarrow { G }^{ 2 }\)
we know, (a + b)² = (a-b)² + 4ab
\(\Rightarrow\) 4A² = (a - b)² + 4G²
\(\Rightarrow\) (a - b)² = 4 (A² -G²)
 \(\Rightarrow\) a-b = 2\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
(1)+(3) \(\rightarrow\)2a = 2\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
\(\Rightarrow\) )a = A+ \(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
(1)-( 3) \(\rightarrow\) 2b = 2\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
\(\Rightarrow\) b = A -\(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)
The Numbers are A\(\pm \) \(\sqrt { { A }^{ 2 }-{ G }^{ 2 } } \)

Let a and r be the first term and common ratio of the G.P.
\(\therefore\)  \({S}_{n}={a(1-r^n)\over1-r},n\epsilon N\)
LHS \(={{S}_{10}-{S}_{20}}^{2}=\left[ {a(1-{r}^{10})\over1-r}-{a(1-{r}^{20})\over1-r} \right]^{2}\)
\(={{a}^{2}\over{{(1-r)}^{2}}}[1-{r}^{10}-1+{r}^{20}]^2\)
\(={{a}^{2}\over{(1-r)}^{2}}.{r}^{20}{({r}^{10}-1)}^{2}={{a^2.{r}^{20}.{({r}^{10}-1)}^{2}}\over{{(1-r)}^{2}}}\)
RHS = S₁₀ (S₃₀ - S₂₀)
\(={a(1-{r}^{10}\over1-r)}\left[ {a(1-{r}^{30})\over1-r}-{{a(1-{r}^{30})}\over{1-r}} \right]\)
\(={{a^2}\over{(1-r^2)}}(1-{r}^{10})[1-{r}^{30}-1+{r}^{20}]={{a^2(1-{r}^{10})}\over{{(1-r)}^{2}}}.{r}^{20}(1-{r}^{10})\)
\(={{a^3.{r}^{20}(1-{r}^{10})^2}\over{{(1-r)}^{2}}}={{{a}^{2}.{r}^{20}{({r}^{10}-1)}^{2}}\over{{(1-r)}^{2}}}\)
\(\therefore\) LHS = RHS.

Let the production in the first year = a units.
Let the annual increase be d units.
\(\therefore\) Production in first year, second year, third year ......... are a, a+d, a+2d, ...
This is an A.P. with first term a and common difference d.
Let T_n be the n^th term of this A.P.
Given T₃ = 600, T7 = 700
\(\therefore\) a + 2d = 600 (1)    and a+6d=700  ....(2)
(2) - (1) \(\rightarrow\) 4d = 100 \(\Rightarrow\) d = 25
\(\therefore\) a + 2 (25) = 600      [ Substituting d = 25 in (1) ]
\(\Rightarrow\)    a + 50 = 600 \(\Rightarrow\) a = 550.
(i) Production in the first year = a = 550 units.
(ii) Total production in 7 years.
= \({S}_{7}={7\over2}[2(550)+6(25)]\left[S_n={n\over 2} \right[2a+(n-1)d]]\)
S₇ = 4375 units
(iii) production in the 10th year = T10 = 550 + 9 (25) = 775 units.

Given T_p+1 = 2.T_q+1
\(\Rightarrow\)   a+(+1-1)d = 2[a+(q+1-1)d]    [ \(\because\) T_n = a + ( n - 1) d ]
\(\Rightarrow\)  a + pd = 2a +2qd
\(\Rightarrow\) a = ( p - 2q ) d   ...(1)
 Now T_3p+1 = a+ ( 3p + 1 - 1 )d = a + 3pd
= ( p - 2q ) d + 3pd   (using (1))
= 4pd - 2qd
= 2d (2p-q) ...(2)
Also T_p+q+1 = a + (p+q+1-1)d
= a+(p+q)d
= (p-2q)d + (p+q)d   (using (1))
= d(p-2q+p+q) = d(2p-q) ...(3)
From (2) and (3), T_3p+1 = 2. T_p+q+1.

Let p = q+h
h is numerically very small and so h₁² h³ ... may be neglected
RHS \(={(n+1)p+(n-1)q\over (n-1)p+(n+1)q}={(n+1)(q+h)+(n-1)q\over (n+1)(q+h)+(n+1)q}\)
\(={nq+q-nh+h+nq-q\over nq-q+nh-h+nq+q}={2nq+(n+1)h\over2nq+(n-1)h}\)
\(={1+{n+1\over2n}.{q\over h}\over1+{n-1\over2n}.{h\over q}}=\left(1+{n+1\over2n}.{h\over q}\right)\left(1+{n-1\over2n}.{h\over q}\right)^{-1}\)
\(=\left({1+{n+1\over2n}.{h\over q}}\right)\left(1-{n-1\over2n}.{h\over q}\right)=1+\left({n+1\over 2n}-{n-1\over2n}\right){h\over q}\)
\(={1+{1\over n}}.{h\over q}\)
LHS \(={p\over q}^{1\over n}=\left(q+h\over q\right)^{1\over n}=\left(1+{h\over q}\right)^{1\over n}=1+{1\over n}.{h\over q}\)
From (1) and (2), LHS = RHS
Now \(\sqrt[8]{15\over16}={(8+1)(15)+(8-1)(16)\over(8-1)(15)+(8+1)(16)}\)  [n = 8, p = 15 and q = 16]
 \(={(9)(15)+7(16)\over(7)(15)+9(16)}={135+112\over105+144}={247\over 249}\)
\(\left( \frac { 15 }{ 16 } \right) =0.9919\)

\(\log { \left( \frac { 1-2x }{ 1+2x } \right) } =log(1-2x)-log(1+2x)\)
\(\log { \left( \frac { 1-2x }{ 1+2x } \right) } =\left[-2x-{(2x)^2\over2}-{(2x)^3\over3}-{(2x)^4\over4}-...\right]-\left[3x-{4x^2\over2}+{8x^3\over3}-{16x^4\over4}+...\right]\)
\(-\left[ 2x-{4x^2\over 2}+{8x^3\over3}-{16x^4\over4}+{32x^5\over 5}+{64x^6\over 6}+{1287x^7\over7}...\right]\)
\(=2\left[-2x-{8x^3\over3}-{32x^3\over5}-{128x^7\over7}-...\right]\)
\(=-2\left[ 2x+{8x^3\over 3}+{32x^5\over 5}+{128x^7\over 7}+...\right]\)
This expression is valid only when \(\left| 2x \right| <1\Rightarrow \left| x \right| <\frac { 1 }{ 2 } \)
This expression is valid in the interval \(-\frac { 1 }{ 2 } \)

\(log\left(1+3\over1-3x\right)=log(1+3x)-log(1-3x)\)[using logarithmic, quotient rule]
\(=\left[ 3x-{(3x)^2\over2}+{(3x)^3\over3}-{(3x)^4\over4}+...\right]-\left[-3x-{(3x)^2\over2}-{(3x)^3\over3}-{(3x)^4\over4}-...\right]\)
\(=\left[ 3x-{9x^2\over2}+{27x^3\over3}-{81x^4\over4}+...\right]-\left[ -3x-{9x^2\over2}-{27x^3\over3}-{81x^4\over4}+...\right]\)
\(=3x-{9x^2\over2}+{27x^3\over3}-{81x^4\over4}x^4+...+3x+{9x^2\over2}+{27x^3\over3}+{81x^4\over4}+...\)
\(=2\left[ 3x+{37x^3\over3}+{243x^5\over5}+{2187x^7\over7}+...\right]\)
This series is valid only when \(\left| 3x \right| <1\Rightarrow \left| x \right| <\frac { 1 }{ 3 } \)
Hence, this series is valid in the interval \(-\frac { 1 }{ 3 }< x<{1\over3}\)

\(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } =\frac { \left( 1+\frac { 2 }{ 3 } (-2x)+..... \right) { \left( 4 \right) }^{ \frac { 3 }{ 2 } }{ \left( 1+\frac { 5 }{ 4 } x \right) }^{ \frac { 3 }{ 2 } } }{ { \left( 1-x \right) }^{ \frac { 1 }{ 2 } } } \)  [using binomial theorem for rational index]
\(={ \left( 1-\frac { 4x }{ 3 } \right) (8) }{ \left( 1+\frac { 3 }{ 2 } \left( \frac { 5 }{ 4 } x \right) \right) }\left( 1-\frac { 1 }{ 2 } (-x) \right) \) [neglecting x², x³ terms....]
\(=8\left( 1-\frac { 4x }{ 3 } \right) \left( 1+\frac { 15x }{ 8 } \right) \left( 1+\frac { x }{ 2 } \right) \)
\(=8\left( 1-\frac { 4x }{ 3 } +\frac { 15x }{ 8 } \right) \left( 1+\frac { x }{ 2 } \right) \)
\(=8\left( 1+\frac { 13x }{ 24 } \right) \left( 1+\frac { x }{ 2 } \right) =8\left( 1+\frac { 13x }{ 24 } +\frac { x }{ 2 } \right) =8\left( 1+\frac { 25x }{ 24 } \right) \)
When x = 0.001, the value of \(\frac { { \left( 1-2x \right) }^{ \frac { 2 }{ 3 } }{ \left( 4+5x \right) }^{ \frac { 3 }{ 2 } } }{ \sqrt { 1-x } } \)
= 8 + \(\frac { 25 }{ 3 } \)(0.001) = 8.01 (upto 2 places )

In (a+b)ⁿ, the general terms is T_r+1 = nC_r a^n-r b^r
To find the Coefficient of 4th term , put r = 3 in (1)
∴ T₄  = nC₃ a^n-3 b³
To find the Coefficient of 13^th them
Put r = 12 in (1)
T₁₃ = nC₁₂ a^n-12 b¹²
Given nC³ = nC₁₂
3 +12  = n
n = 15

General term in (1+x)ⁿ is
T_r+1 = nC_r(1)^n-r.x^r = nC_rx^r
\(\therefore\) Co-efficient of T_r+1 is n C_r.
Let the Co-efficients of T_r+1, T_r+2 and T_r+3 be in the ratio 1 : 3 : 5
\(\therefore\) nC_r : nC_r+1 : nC_r+2 = 1: 3: 5
\(\Rightarrow\)  \({{n{C}_{n1}}\over{n{C}_{r}}}={3\over 1}\)   ....(1) 
and \({n{C}_{n+2}\over{C}_{n+1}}={5\over3}\)    ...(2)
From (1)  \(\rightarrow{\frac { \frac { n! }{ (r+1)!(n-r-1)! } }{ \frac { n! }{ r!(n-r)! } } }=3\)
\(\Rightarrow\)   \(\frac{r!(n-r)!}{(r+1)!(n-r-1)!}=3\)
\(\Rightarrow\)      
\(\Rightarrow{n-r\over+1}=3\)
Similarly (1) implies
\(\frac { \frac { n! }{ (r+2)!(n-r-2)! } }{ \frac { n! }{(r+1)!(n-r-1)! } } =\frac{5}{3}\)
\(\frac{(r+1)!(n-r-1)!}{(r+2)!(n-r-2)!}=\frac{5}{3}\Rightarrow\)  
\(\Rightarrow{n-r-1 \over r+2}={5\over 3}\Rightarrow 3n-3t-3=5r+10\)
\(\Rightarrow\)  3n - 8r - 13 = 0
(3) X 2 \(\rightarrow\) 2n-8r-6 = 0
 

We have (x + y) n = nC₀xⁿy⁰ + nC₁x^n-1y¹ + nC₂x^n-2y²+....+ nC_nx⁰y²
= (nC₀ xⁿy⁰ + nC₂ x^n-2y²+...) + (nC₁ x^n-1y + nC₃x^n-3y³+....)    [Separated the odd and even terms]
(x + y) n = P + Q  ...(1)      [ \(\because\) P in the sum of odd terms and Q is the sum of even terms ]
Also (x-y)n = nC₀xⁿy⁰- nC₁x^n-1y¹ + nC₁x^n-1y¹ + nC₂x^n-2y²+......+ nC_nx⁰(-y)ⁿ
= (nC₀xⁿy⁰+nC₂x^n-2y²+.3...)-(nC₁x_n-1y¹+nC₃x^n-3y³+.....)...(2)
(x - y)n = P-Q
(i) Multipying (1) and (2) we get,
(x+y)ⁿ(x-y)ⁿ = (P+Q)(P-Q)
\(\Rightarrow\) [(x+y)(x-y)]n = P²-Q²
\(\Rightarrow\)(x²-y²)n = P²- Q². Hence proved..
(ii) From (1) (x+y)2n = (P+Q)² = P² + Q² + 2PQ ....(3)
From (2) (x-y)2n = (P-Q)2 = P²+Q²-2PQ  ....(4)
Subtracting (4) from (3) we get,
(x+y)²ⁿ - (x-y)2n = P²+Q²+2PQ-(P²+Q²-2PQ)
= P² + Q² + 2PQ - P² - Q² + 2PQ
= 4PQ.

Let a = a +b -b = b+ (a-b)
aⁿ  = [b + (a-b)]ⁿ
Using binomial theoream
aⁿ = bⁿ + nC₁ b^n-1 (a-b) + nC₂ b^n-2 (a-b)² + ...+ (a-b)ⁿ
aⁿ-bⁿ  = nC₁ b^n-1 (a-b) + nC₂ (a-b)² b^n-2 +...+(a-b)ⁿ
aⁿ-bⁿ = (a-b) [nC₁ b^n-1 + nC₂ b^n-2 (a-b) +...+ (a-b)^n-1]
aⁿ-bⁿ = (a-b) K [where k = nC₁ b^n-1 +nC₂ b^n-2 (a-b) +...+(a-b)n^-1
Thus, (a - b) is a factor of aⁿ - bⁿ

We have (1 +x)ⁿ = nCo(1)ⁿ + nC₁(1)^n-1 . x¹ + nC₂(1)^n-2 x² + ...+ nC_nxⁿ;
⇒ (1 +x)ⁿ = C₀ + C₁x + C₂x₂+ ...+ Cn_nxⁿ  ...(i)
Similarly, (x + 1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-1 + ...+ Cn
Multiplying equations (1) and (2) we get
(1 + x)ⁿ (x + 1)ⁿ = [Co + C₁x + C₂x² + ...+C_nxⁿ]  [C₀xⁿ + C₁x^n-1 + C₂x^n-2+...+ C_n]
(1 +x)²ⁿ = [C₀ + C₁x + C₂x²+ ... +C_nxⁿ][C₀xⁿ + C₁x^n-1 + C₂x^n-2 +...+ C_n]     ...(3)
This relation is true for all values of n. Let us equate the co-efficients of xⁿ both sides​​​​​​​
General term in (1 + x)²ⁿ is T_r+1 = 2nC_r(1)^2n-rx^r​​​​​​​
Putting r = n we get​​​​​​​
T_n+1= 2_nC_n.xⁿ
⇒ Co-efficient of xⁿ of LHS of(3) is 2_nC_n
Now, co-efficient of xⁿ in RHS of (3) is​​​​​​​
\(C_0^2+C_1^2+C_2^2+...+C_n^2\)
\(∴\ C_0^2+C_1^2+C_2^2+...+C_n^2=2_nC_n\)
\(⇒\ C_0^2+C_1^2+C_2^2+...+C_n^2={(2n)!\over n!n!}\left[ ∵ nC_r={n!\over r!(n-r)!}\right]\)
\(={(2n)!\over (n!)^2}\)
Hence \({ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+...C_{ n }^{ 2 }=\frac { (2n)! }{ (n)! } \)

In (1 + x)ⁿ, the general term is
T_r+1 = nCr (1)^n-r x^r ....(1)
To find the co-efficients of 5^th... , 6^th .. and 7^th ... terms, put r = 4, 5, 6 in (1) respectively.
∴Coefficients of 5^th ... , 6^th .... and 7^th ... terms in (1 +x)ⁿ are nC₄, nC₅ and nC₆ respectively.
Given that nC₄, nC₅ and nC₆ are in A.P
\(∴\ 2nC_5=nC_4+nC_6\)
\(⇒\ 2={nC_4\over nC_5}+{nC_6\over nC_5}\)    \(\left[∵{nC_r\over nC_{r-1}}={n-r+1\over r}\right]\)
\(⇒ 2={5\over n-4}+{n-5\over 6}\) 
\(⇒\ 12={30+(n-5)(n-4)\over 6(n-4)}\)

⇒ 12(n - 4) = 30 + n² - 4n - 5n + 20
⇒ 12n - 48 = n² - 9n + 50
⇒ n² - 9n + 50 - 12n + 48 = 0
⇒ n² - 21n + 98 = 0
⇒ (n - 14)(n - 7) = 0
⇒ n = 7 or 14.