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

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

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

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

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

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

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

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

\(\sqrt [ 3 ]{ x^3+7 } ={(x^3+7)}^{{1\over 3}}\)
\(={\left[ x^3\left( 1+{7\over x^3} \right) \right]}^{{1\over 3}}\)   (\(\left |{7\over x^3}\right |<1\) 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
\(\sqrt [ 3 ]{ x^3+7 } -\sqrt [ 3 ]{x^3+4 } =\left(x+{7\over 3}\times{1\over x^2} \right)-\left( x+{4\over 3}\times{1\over x^2} \right)={1\over x^2}\)

According to the given information, we have the following diagram.

Distance travelled to bring first ball = 24 + 24 = 2 \(\times\) 24 = 48 m
Distance travelled to bring second ball = 2 (24 + 4) = 2(28) = 56 m
Distance travelled to bring third ball = 2 (24 + 4 + 4) = 2(32) = 64 m
\(\therefore\) The series of distances are 48, 56, 64 ...
Here a = 48, d = 56 - 48 = 8 and n = 20.
To find the total distance that he run in bringing back all balls, we have to find the sum of 20 terms of the above series
\(\therefore\) \({ S }_{ 20 }=\frac { 20 }{ 2 } \left[ 2\left( 48 \right) +19\left( 8 \right) \right] \)
= 10[96 + 152]
= 10[248]
S₂₀ = 2480 m.

Given equation is (q - r).x²+ (r - p)x +p - q = 0
a = q - r, b = r - p, c = p - q
Since the roots of the quadratic equation are equal, b² - 4ac = 0
\(\Rightarrow\) (r - p)² - 4(q - r)(p - q) = 0
\(\Rightarrow\) r² +p² - 2rp - 4(pq - q² - rp + rq) = 0
\(\Rightarrow\) r² + p² - 2rp - 4pq + 4q² + 4rp - 4rq = 0
\(\Rightarrow\) r² + p² + 4q² + 2rp - 4pq - 4rq = 0
\(\Rightarrow\) (r + p - 2q)² = 0
\(\Rightarrow\) r + p - 2q = 0
\(\Rightarrow\) 2q = r + p
\(\Rightarrow\) q - p = r - p
\(\Rightarrow\) common difference is equal for p, q, r
Henc p, q, r are in A.P.

\(\sqrt [ 3 ]{ 126 } ={ (125) }^{ 1/3 }=(125+1)^{ 1/3 }=\left\{ 125\left( 1+\frac { 1 }{ 125 } \right) \right\} ^{ 1/3 }=(125)^{ 1/3 }\left[ 1+\frac { 1 }{ 125 } \right] ^{ 1/3 }\)
\(=5\left[ 1+\frac { 1 }{ 3 } \times \frac { 1 }{ 125 } +... \right] \left( \therefore \frac { 1 }{ 125 } <1 \right) =5\left[ 1+\frac { 1 }{ 3 } (0.008) \right] =5(1+0.002666)=5.01\)

In (1+2x)⁸ , we have n = 8, x = 1, a = 2x.
T_r+1 = nC_r x^n-r a^r
⇒ T_r+1 = 8Cr(1)^8-r.(2x)^r = 8Cr 2^r.x^r  --- (1)
and T_r-1 = 8C_r-1.(2x)^r-1 = 8C_r-1.2^r-1x^r-1 ----- (2)
Dividing (2) ÷ (1) we get,
\(\frac{T_{r+1}}{T_r}=\frac{8C_r.2^{r}.x^{r}}{8C_{r-1}.2^{r-1}.x_{r-1}}=\frac{8!}{r!(8-r)!}.\frac{(r-1)!(8-r+1)}{8!}2x\)
= \(\frac{8-r+1}{r}.2r=\frac{9-r}{r}.2(2)\)   [since x=2]
= \(\frac{36-4r}{r}\)
Now T_r+1 ≥T _r if \(\frac{T_{r+1}}{T_{r}}\ge1\)
⇒ \(\frac{36-4r}{r}\ge1\)
⇒ 36 - 4r≥1
⇒ 5r ≤ 36
⇒ r ≤ \(\frac{36}{5}\)
⇒ r ≤ 7.2
∴ the greatest possible value of r is 7.
The greatest possible value of r is 7.

Given \({ \left( x-\frac { 1 }{ 2y } \right) }^{ 10 }\)
Here n = 10, x = x and \(a=\left( \frac { -1 }{ 2y } \right) \)
Middle term = \({ T }_{ \frac { 10+2 }{ 2 } }={ T }_{ 6 }\)
General term is \({ T }_{ r+1 }=nCr{ x }^{ n-r }{ a }^{ r }\)
Putting r = 5 we get,
\({ T }_{ 6 }=10{ C }_{ 5 }{ x }^{ 10-5 }{ \left[ -\frac { 1 }{ 2y } \right] }^{ 5 }=\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } .{ x }^{ 5 }\left( \frac { -1 }{ 32.{ y }^{ 5 } } \right) \)
\(=-225.{ x }^{ 5 }.\frac { 1 }{ 32{ y }^{ 5 } } { T }_{ 6 }=\frac { -63{ x }^{ 5 } }{ 8{ y }^{ 5 } } \)

Given \({ \left( \frac { 4x }{ 5 } -\frac { 5 }{ 2x } \right) }^{ 9 }\)
Here n = 9, x = \(\frac{4x}{5}\) and a = \((\frac{-5}{2x})\)
\(\therefore { T }_{ r+1 }={ 9C }_{ r }{ \left( \frac { 4x }{ 5 } \right) }^{ 9-r }{ \left( \frac { -5 }{ 2x } \right) }^{ r }\)
\(={ 9C }_{ r }.\frac { { 4 }^{ 9-r } }{ { 5 }^{ 9-r } } .{ x }^{ 9-r }{ \left( -1 \right) }^{ r }.\frac { { 5 }^{ r } }{ { 2 }^{ r }.{ x }^{ r } } \)
\(={ \left( -1 \right) }^{ r }9Cr\frac { { 12 }^{ 18-3r } }{ { 5 }^{ 9-2r } } .\frac { { 5 }^{ r } }{ { 2 }^{ r } } .{ x }^{ 9-2r }\)
\({ T }_{ r+1 }={ \left( -1 \right) }^{ r }9Cr\frac { { 12 }^{ 18-3r } }{ { 5 }^{ 9-2r } } .{ x }^{ 9-2r },0\le r\le 9.\)

\((1+x)^{ m }=1+mx+\frac { m(m-1) }{ 2! } { x }^{ 2 }+..\)    [Binomial theorem for rational index]
∴ Co-efficient of x² = \(\frac { m(m-1) }{ 2 } \)
Given \(\frac { m(m-1) }{ 2 } \) = 6 ⇒ m²- m = 12
⇒ m²-m -12 = 0 ⇒ (m - 4) (m + 3) = 0
⇒ m = 4 or -3
∴ Negative value of m is -3

Given series is  1² + 3² + 5² +...
Let Tn be the nth term
T_n = (nth term of 1, 3, 5,...)²
= [1+(n-1)2]² = (1 + 2n - 2)² = (2n-1)²
= 4n² + 1 - 4n
∴ Sum of n terms =  \(\sum { 4{ n }^{ 2 } } -4n+1=4\sum { n^{ 2 } } -4\sum { n } +n\)
= \(4\frac { (n)(n+1)(2n+1) }{ 6 } -\frac { 4n(n+1)+n }{ 2 } \)
= \(\frac { n }{ 2 } \) [2(n + 1)(n + 1) - 6(n + 1) + 3]
= \(\frac { n(4{ n }^{ 2 }-1) }{ 3 } \)

Consider the terms in the numerator 1, 2, 3....
a = 1, d = 2 -1 = 1 a_n = a + (n-1) d
a_n = 1 + (n-1) (1) = 1 + n - 1 = n
The terms in the denominator are 2, 3, 4, 5, 6...
here  a = 2, d = 1
a_n = 2 + (n-1) 1 = 2 + n -1 = n + 1
Hence n^th term of the given sequence is \(\frac { n }{ n+1 } \)
 

2018
Let a_n = 2018
then the first 6 terms are 2018, 2018, 2018, 2018, 2018, 2018
It is not an AP, GP, AGP and HP.

Let a_n = \(\frac { 2n+3 }{ 3n+4 } \)
\({ a }_{ 1 }=\frac { 2+3 }{ 3+4 } =\frac { 5 }{ 9 } \)
\({ a }_{ 2 }=\frac { 4+3 }{ 6+4 } =\frac { 7 }{ 10 } \)
\({ a }_{ 3 }=\frac { 6+3 }{ 9+4 } =\frac { 9 }{ 13 } \)
\({ a }4=\frac { 8+3 }{ 12+4 } =\frac { 11 }{ 16 } \)
\({ a }_{ 5 }=\frac { 10+3 }{ 15+4 } =\frac { 13 }{ 19 } \)
\({ a }_{ 6 }=\frac { 12+3 }{ 18+4 } =\frac { 15 }{ 22 } \)
\(\frac { 5 }{ 9 } ,\frac { 7 }{ 10 } ,\frac { 9 }{ 13 } ,\frac { 11 }{ 16 } ,\frac { 13 }{ 19 } ,\frac { 15 }{ 22 } ...\)
this is neither A.P, G.P nor AGP

Let \(a_n={(-1)^n\over 1}\)
\(a_1={(-1)^1\over 1}=-1, a_2={(-1)^2\over 2}={1\over 2}, a_3={(-1)^3\over 3}={-1\over 3}\)
\(a_4 ={(-1)^4\over 4}={1\over 4},a_5 ={(-1)^5\over 5}=-{1\over 5},a_6 ={(-1)^6\over 6}={1\over 6}\)
∴ The sequence is  \(-1,{1\over 2},-{1\over 3},{12\over4},-{1\over 5}, {1\over 6},...\)
That is \(-1,{1\over 2},-{1\over 3},{12\over4},-{1\over 5}, {1\over 6},...\)
Consider 1, 2, 3, 4, .... which is an A.P.
Since d = 2 - 1 = 3 - 2 = 1
and -1, 1, -1,1, ... is a G.P. where
\(r=\frac { 1 }{ -1 } =\frac { -1 }{ 1 } =-1\)
Hence this is an arithmetico-geometric progression.

We have \({\left( 2x-{1\over 2x} \right)}^{4}\) = ⁴C₀(2x)⁴ \({\left(-{1\over 2x} \right)}^{0}\) +⁴C₁(2x)³\({\left(-{1\over 2x} \right)}^{1}\) + ⁴C₂(2x)²\({\left(-{1\over 2x} \right)}^{2}\)+⁴C₃(2x)¹\({\left(-{2\over x} \right)}^{3}\) + ⁴C₄(2x)⁰\({\left(-{1\over 2x} \right)}^{4}\)
= (2x)⁴ - 4(2x)³\({\left({1\over 2x} \right)}\) + 6(2x)²\({\left({1\over 2x} \right)}^{2}\)- 6(2x)\({\left({1\over 2x} \right)}^{2}+{\left({1\over 2x} \right)}^{4}\) 
\(=16x^4-16x^2+6-{3\over2x^2}+{1\over16x^4}\)