LHS = \(\sqrt { \frac { 1-x }{ 1+x } } \)
\(\sqrt { \frac { 1-x }{ 1+x } } =\sqrt{(1-x)(1-x)\over(1+x)(1-x)}\)
\(={1-x\over\sqrt{1-x^2}}\)
\(=(1-x)(1-x^2)^{-1\over2}\)
\(=(1-x)\left[ 1+{-1\over2}(-x^2)-{\left(-1\over2\right)\left({-1\over2}-1\right)\over2.1}(-x^2)+... \right]\)∵ x is very small, [x² is also very small]
\(=(1-x)\left[ 1+{x^2\over2}+{\left(1\over2\right)\left(3\over2\right)\over1.2}(x^4)+... \right]\)
\(=(1-x)\left[1+{x^2\over2}+{3\over8}x^4+...\right]=1-x+{x^2\over2}-{x^2\over2}+{3\over8}x^4+...\)
\(\left( 1-x+\frac { { x }^{ 2 } }{ 2 } \right) \) approximately
RHS.
Hence proved.

In \(\left( { 2x }^{ 3 }-\frac { 1 }{ { 3x }^{ 2 } } \right) ^{ 5 }.n = 5, x = 2x^3, a=-{1\over 3x^2}\)
∴ General term is
T_r+1 = nC_r x^n-ra^r
\(=5C_r (2x^3)^{5-r}\left(-{1\over 3x^2}\right)^r\)
\(= 5C_r 2^{5-r}x^{15-3r}{(-1)^r\over 3^r.x^{2r}}\)
\(=5C_r{2^{5-r}\over 3^r}(-1)^rx^{15-3r-2r}\)
To get the constant term, put 15 - 5r = 0
⇒ 15-5r = 0
⇒ 15 = 5r ⇒ \(r={15\over 5}=3\)
Putting r = 3 in (1) we get
\(T_4=5C_3{2^2\over 3^3}.(-1)^3.x^0\)
\(=-5C_3{(4)\over 27}\)
\(=-{5\times4\times3\over 3\times2\times1}\times{4\over 27}\)
= \(-\frac { 40 }{ 27 } \)
Hence the constant term is \(-\frac { 40 }{ 27 } \)

Clearly, number of bacteria at the end of different hours forms a G.P. with
a = 30 and r = 2.
Number of bacteria present at the end of 2^nd hour
t₃ = a . r² = 30 \(\times\) 2² = 30 \(\times\) 4 = 120.
Number of bacteria present at the end of 4^th hour
t₅ = a . r⁴ = 30 (2⁴) = 30 (16) = 480.
Number of bacteria present at the end of n^th hour
t_n+1 = a . rⁿ = 30 (2ⁿ).

Let T_n be the nth term of the given series
Then T_n = 1 + 4 + 4² + 4³ + ...
\(=1\left(4^n-1\over 4-1\right)\)
\(={4^n-1\over 3}\)
Let S_n be the sum to n terms of the given series
Then \(S_n={\sum_{k=1}^n}T_k=\sum_{k=1}^n{4^n-3\over3}\)
\(⇒\ S_n={1\over3}\left[ \sum_{k=1}^n4^n-\sum_{k=1}^n3\right]\)
\(⇒\ S_n= {1\over3}[4^1+4^]+...+4^n-3^n\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]={1\over 3}\left[4{(4^n-1)-9n\over3}\right]\)

\({ S }_{ n }=\frac { 4 }{ 9 } \left[ \left( { 4 }^{ n }-1 \right) -n/3 \right] \)

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 (1 + 11 + 111 + 1111 + ....) upto n terms
\(={8\over9}(9 + 99 + 999 + ...)\)
\({ S }_{ n }=\frac { 8 }{ 81 } \left[ \left( { 10 }^{ n }-1 \right) -9n \right] \) [multiplying and dividing by 9]
\(={8\over9}[10 -1) + (100 -1) + (1000 -1) + ...]\)
\(S_n={8\over 9}[(10^1 +10^2 +10^3 + ... +10^n)-(1+1+1+ ... +1n\ terms)]\)
In 10 + 10² + 10³ + ... + 10ⁿ, a = 10, r= 10, and it forms a G.P.
\(∴\ S_n={a(r^n-1)\over r-1}=10{(10^n-1)\over 10-1}={10\over 9}(10^n)-1\) and 1 + 1 + 1 ... + upto n terms = n
Substituting these values in (1) we get
\(S_n={8\over 9}\left[ 10(10^n-1)n\over 9\right]\)
\(S_n={8\over 81}[(10^n-1)-9n]\)

Let T_n be the n^th term of the given series
\(T_n={1^3+2^3+..+n^3\over 1+3+5+...+(2n-1)}={\left[n(n+1)\over\right]^2\over{n\over2}(1+2n-n)}\ \ \left[ ∵\ S_n={n\over2}(a+1)\right]\)
\(={n^2(n+1)^2\over4}/{n\over2}(2n)\)
\(={n^2(n+1)^2\over4}\times{1\over n^2}={(n+1)^2\over2}\)
\(={1\over2}(n^2+2n+1)\)
Let S_n denote the sum of n terms of the given series
Then \(S_n=\sum _{k=1}^nT_k={1\over 4}(k^2+2k+1)\)
\(={1\over4}\left[ \sum_{k=1}^nK^2+2\sum_{k=1}^nk+\sum_{k=1}^n1\right]\)
\(={1\over24}[n(n + 1)(2n + 1)+ 6(n)(n + 1)+ 6n]\)
\(={1\over24}[(n^2 +n)(2n+1)+6n^2 +6n+6n]\)
\(={1\over24}[2n^3 + n^2 + 2n^2 + n + 6n^2 + 12n]\)
\(S_n={1\over24}[2n^3 +9n^2 +13n]={n\over 24}[2n^2+9n+13]]\)
Now we have to find S₁₇
\(∴ S={17\over24}[2(17)^2 + 9(17)+ 13]={17\over 24}[578+153+13]\)
\(={17\over 24}(744)=17(31)=527\)
S₁₇ = 527

= [(x-a)ⁿ = xⁿ + ⁿC₁x^n-1(-a)¹+ⁿC₂x^n-1(-a)²+.....(-a)ⁿ]
= \(\left[ \left( { 2x }^{ 2 } \right) ^{ 4 }-4C_{ 1 }\left( 2x^{ 2 } \right) \left( 3\sqrt { 1-{ x }^{ 2 } } \right) ^{ 2 }+4C_{ 2 }\left( { 2x }^{ 2 } \right) (3\sqrt { 1-{ x }^{ 2 }) } ^{ 2 }-4C_{ 3 }({ 2x }^{ 2 })^{ 1 }(3\sqrt { 1-{ x }^{ 2 } } )^{ 3 }+(3\sqrt { 1-{ x }^{ 2 } } )^{ 4 } \right] \)  \(=\left[ \left( { 2x }^{ 2 } \right) ^{ 4 }-4C_{ 1 }\left( 2x^{ 2 } \right) ^{ 3 }\left( 3\sqrt { 1-{ x }^{ 2 } } \right) ^{ 1 }+4C_{ 2 }\left( { 2x }^{ 2 } \right) (3\sqrt { 1-{ x }^{ 2 }) } ^{ 2 }+4C_{ 3 }({ 2x }^{ 2 })(3\sqrt { 1-{ x }^{ 2 } } )^{ 3 }+(3\sqrt { 1-{ x }^{ 2 }) } ^{ 4 } \right] \)= \(2\left[ \left( { 2x }^{ 2 } \right) ^{ 4 }+4C\left( { 2x }^{ 2 } \right) ^{ 2 }(3\sqrt { 1-{ x }^{ 2 } } )^{ 2 }+(3\sqrt { 1-{ x }^{ 2 } } )^{ 4 } \right] \)
= \(2\left[ \left( 16{ x }^{ 8 } \right) +\frac { 4\times 3 }{ 2\times 1 } \times { 4x }^{ 4 }\times 9(1-{ x }^{ 2 })+{ 3 }^{ 4 }(1-{ x }^{ 2 })^{ 2 } \right] \)
= \(2\left[ 16{ x }^{ 8 }+216{ x }^{ 4 }(1-{ x }^{ 2 })+81(1-{ x }^{ 2 })^{ 2 } \right] \)

Let A be the first term and R be the common ratio of the given G.P.
Then a = p^th term ⇒ a = ARP^p-1
⇒ log a log A+(p-1 )logR...(1)
b = q^th term b = AR^q-1
⇒ log b = logA +(q-1) log R...(2)
c = r^th term ⇒ c = AR^r-1
⇒ log c = log A + (r-1) log R
Now, LHS = (q - r) log a + (r - p) log b + (p - q) log c
= (q - r) [log A + (p - 1) log R] + (r - p) [log A + (q - 1) log R] + (p - q)[log A + (r-1) log R]
= log A [q - r + r - p + P - q] + log R [(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)]
= log A (0) + log R [pq - pr - q + r + qr - pq - r + p + rp - rq - p + q]
= log R [0] = 0
∴ (q - r) log a + (r- p) log b + (p - q) log c = 0.

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.

Given \(\sqrt [ 3 ]{ 1001 } ={ \left( 1000+1 \right) }^{ \frac { 1 }{ 3 } }={ \left( 1000 \right) }^{ \frac { 1 }{ 3 } }{ \left( 1+\frac { 1 }{ 1000 } \right) }^{ \frac { 1 }{ 3 } }\)
\(={ 10 }^{ 3\times \frac { 1 }{ 3 } }{ \left[ 1+\frac { 1 }{ 1000 } \right] }^{ \frac { 1 }{ 3 } }\)
\(\sqrt [ 3 ]{ 1001 } =10{ \left( 1+.001 \right) }^{ \frac { 1 }{ 3 } }\)
\(=10\left[ 1+\frac { .001 }{ 3 } +\left( \frac { 1 }{ 3 } \right) \left( -\frac { 2 }{ 3 } \right) \left( \frac { .000001 }{ 2 } \right) \right] app\)
\(=10\left[ 1+.00033-\frac { .000001 }{ 9 } \right] app\)
= 10 [1.00033 - .00000011] app
= 10 [1.000329]
= 10 [1.00033]
\(\\ \\ { \left( 1000 \right) }^{ \frac { 1 }{ 3 } }\cong 10.0033\)