By using Binomial Theorem, the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\) can be expanded as,
\(\left(\frac{x}{3}+\frac{1}{x}\right)^{5} ={ }^{5} C_{0}\left(\frac{x}{3}\right)^{5}+{ }^{5} C_{1}\left(\frac{x}{3}\right)^{4}\left(\frac{1}{x}\right)+{ }^{5} C_{2}\left(\frac{x}{3}\right)^{3}\left(\frac{1}{x}\right)^{2} \)
\(+{ }^{5} C_{3}\left(\frac{x}{3}\right)^{2}\left(\frac{1}{x}\right)^{3}+{ }^{5} C_{4}\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^{4}+\xi_{5}\left(\frac{1}{x}\right)^{5} \)
\(=\frac{x^{5}}{243}+5\left(\frac{x^{4}}{81}\right)\left(\frac{1}{x}\right)+10\left(\frac{x^{3}}{27}\right)\left(\frac{1}{x^{2}}\right)+10\left(\frac{x^{2}}{9}\right)\left(\frac{1}{x^{3}}\right)+5\left(\frac{x}{3}\right)\left(\frac{1}{x^{4}}\right)+\frac{1}{x^{5}} \)
\(=\frac{x^{5}}{243}+\frac{5 x^{3}}{81}+\frac{10 x}{27}+\frac{10}{9 x}+\frac{5}{3 x^{3}}+\frac{1}{x^{5}}\)

96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.
It can be written that, 96 = 100 – 4
\(\left( 96 \right) ^{ 3 }=\left( 100-4 \right) ^{ 3 }\)
\(=^{ 3 }{ C }_{ 0 }\left( 100 \right) ^{ 3 }-^{ 3 }{ C }_{ 1 }\left( 1000 \right) ^{ 2 }\left( 4 \right) ^{ 1 }+^{ 3 }{ C }_{ 2 }\left( 100 \right) ^{ 1 }4^{ 2 }+^{ 3 }{ C }_{ 3 }4^{ 3 }\)
\( =884736\)

By using Binomial Theorem, the expression (1– 2x)⁵ can be expanded as,
\((1-2x)^{ 5 }=\left[ 1+\left( -2x \right) \right] ^{ 5 }\)
\(=^{ 5 }{ C }_{ o }+^{ 5 }{ C }_{ 1 }(-2x)+^{ 5 }{ C }_{ 2 }(-2x)^{ 2 }+^{ 5 }{ C }_{ 3 }(-2x)^{ 3 }+^{ 5 }{ C }_{ 4 }(-2x)^{ 4 }+^{ 5 }{ C }_{ 5 }(-2x)^{ 5 }\)
\(=1+5(-2x)+10({ 4x }^{ 2 })+10({ -8x }^{ 3 })+5({ 16x }^{ 4 })+1(-{ 32x }^{ 5 })\)
\(=1-10x+{ 40x }^{ 2 }-{ 80x }^{ 3 }+{ 80x }^{ 4 }-{ 32x }^{ 5 }\)
which is the required expansion.

Use(1+x)^{n = n}C_o+ ⁿC₁x + ⁿC₂x^{2 + n}C₃x³+......+ⁿC_nxⁿ
We have, \(\overset { n }{ \underset { r=0 }{ \Sigma } } \quad ^{ n }{ C }_{ r }\times { 3 }^{ r }=^{ n }{ C }_{ o }{ 3 }^{ o }+^{ n }{ C }_{ 1 }{ 3 }+^{ n }{ C }_{ 2 }{ 3 }^{ 2 }+^{ n }{ C }_{ 3 }{ 3 }^{ 3 }+...+^{ n }{ C }_{ n }{ 3 }^{ n }\)  [on putting r=0, 1, 2,....,n]
\(^{ n }{ C }_{ o }+^{ n }{ C }_{ 1 }3+^{ n }{ C }_{ 2 }{ 3 }^{ 2 }+^{ n }{ C }_{ 3 }{ 3 }^{ 3 }+...+^{ n }{ C }_{ n }{ 3 }^{ n }\)
\(=(1+3)^{ n }\quad \left[ \because \left( 1+x \right) ^{ n }=^{ n }{ C }_{ o }+^{ n }{ C }_{ 1 }x+^{ n }{ C }_{ 2 }{ x }^{ 2 }+...+^{ n }{ C }_{ n }{ x }^{ n } \right] \)
= 4ⁿ

Splitting 1.01 and using binomial theorem to write the first few terms we have
(1.01)^1000000 = (1 + 0.01)^1000000
= ^1000000C₀ + ^1000000C₁(0.01) + other positive terms
= 1 + 1000000 x 0.01 + other positive terms
= 1 + 10000 + other positive terms
> 10000
Hence (1.01)^1000000 > 10000.

By using binomial theorem, we have
\(\left(x^{2}+\frac{3}{x}\right)^{4} ={ }^{4} \mathrm{C}_{0}\left(x^{2}\right)^{4}+{ }^{4} \mathrm{C}_{1}\left(x^{2}\right)^{3}\left(\frac{3}{x}\right)+{ }^{4} \mathrm{C}_{2}\left(x^{2}\right)^{2}\left(\frac{3}{x}\right)^{2}+{ }^{4} \mathrm{C}_{3}\left(x^{2}\right)\left(\frac{3}{x}\right)^{3}+{ }^{4} \mathrm{C}_{4}\left(\frac{3}{x}\right)^{4}\)
\(=x^{8}+4 x^{6} \cdot \frac{3}{x}+6 \cdot x^{4} \cdot \frac{9}{x^{2}}+4 \cdot x^{2} \cdot \frac{27}{x^{3}}+\frac{81}{x^{4}} \)
\(=x^{8}+12 x^{5}+54 x^{2}+\frac{108}{x}+\frac{81}{x^{4}}\)

Middle term in \({ \left( 1+\alpha x \right) }^{ 4 }\) = \(\left( \frac { 4 }{ 2 } +1 \right) \)th = 3rd term
\(\therefore \) Coefficient of middle term = ⁴C₂ (\(\alpha \))²
Middle term in \({ \left( 1-\alpha x \right) }^{ 6 }\)= \(\left( \frac { 6 }{ 2 } +1 \right) \)th = 4th term
\(\therefore \) Coefficient of middle term = (-1)^3 6C₃ (\(\alpha \))³
Now, ⁴C₂ (\(\alpha \))² = - ⁶C₃ (\(\alpha \))³
\(\Rightarrow 6{ \alpha }^{ 2 }+20\alpha ^{ 3 }=0\Rightarrow 2{ \alpha }^{ 2 }\left( 3+10\alpha \right) =0\)
\( \Rightarrow \alpha =0,\frac { -3 }{ 10 } \)

\(\left( 12 \right) ^{ 5 }+\left( 8 \right) ^{ 5 }=\left( 10+2 \right) ^{ 5 }+\left( 10-2 \right) ^{ 5 }\\ \left( 10+2 \right) ^{ 5 }=^{ 5 }{ C }_{ 0 }\times { 10 }^{ 5 }+^{ 5 }{ C }_{ 1 }\times { 10 }^{ 4 }\times 2+^{ 5 }{ C }_{ 2 }\times { 10 }^{ 3 }\times { 2 }^{ 2 }+.........(i)\\ \left( 10-2 \right) ^{ 5 }=^{ 10 }{ C }_{ 5 }\times 5-^{ 5 }{ C }_{ 1 }\times { 10 }^{ 4 }\times 2+^{ 5 }{ C }_{ 2 }\times { 10 }^{ 3 }\times { 2 }^{ 2 }+.........(ii)\\ Now\quad adding\quad equ(i)\quad and\quad (ii)\\ \left( 12 \right) ^{ 5 }+\left( 8 \right) ^{ 5 }=2\left[ ^{ 5 }{ C }_{ 0 }10^{ 5 }+^{ 5 }{ C }_{ 2 }\left( 10 \right) ^{ 3 }\left( 2 \right) ^{ 2 }+^{ 5 }{ C }_{ 4 }\left( 10 \right) ^{ 1 }\left( 2 \right) ^{ 4 } \right] \\ =281600\)

Here n = 9 which is odd
So the middle terms are \(\left( \frac { 9+1 }{ 2 } \right) and\left( \frac { 79+1 }{ 2 } +1 \right) \) th i.e. 5^th and 6^th terms.
The general term in the expansion of \(\left( 2x-\frac { { x }^{ 2 } }{ 6 } \right) \)is
\({ T }_{ r+1 }=^{ 9 }{ C }_{ r }(2x)^{ a-r }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }\)
Putting r = 4 and 5 in (i)
\({ T }_{ 5 }=^{ 9 }{ C }_{ 4 }(2x)^{ 9-4 }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }=^{ 9 }{ C }_{ 4 }(2x)^{ 5 }(-1)^{ 4 }\left( \frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }\)
\(=\frac { 9! }{ 4!5! } \times 32{ x }^{ 5 }\times \frac { { x }^{ 8 } }{ 1296 } =\frac { 9\times 8\times 7\times 6 }{ 4\times 3\times 2\times 1 } \times 32{ x }^{ 5 }\times \frac { { x }^{ 8 } }{ 1296 } =\frac { 28 }{ 9 } x^{ 13 }\)
\({ T }_{ 6 }=^{ 9 }{ C }_{ 5 }(2x)^{ 9-5 }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 5 }=^{ 9 }{ C }_{ 5 }(2x)^{ 4 }(-1)^{ 5 }\left( \frac { { x }^{ 2 } }{ 6 } \right) ^{ 5 }\)
\(=\frac { 9! }{ 5!4! } \times 16{ x }^{ 4 }\times \frac { { x }^{ 10 } }{ 7776 } =-\frac { 7 }{ 27 } { x }^{ 14 }\) 

We have (3x² - 2ax + 3a²)³
=[(3x²- 2ax) + 3a²)]³
=³C₀(3x²- 2ax)³ + ³C₁(3x² - 2ax)² (3a²)+³C₂(3x²- 2ax) (3a²)²+ ³C₃(3a²)³
= (3x² - 2ax)³ + 3 x 3a² (3x² - 2ax)² + 3x 9a⁴ (3x² - 2ax) + 27a⁶
= (27x⁶ - 8a³x³ - 54ax⁵ + 36a²x⁴) + 9a² (9x⁴+ 4a²x² - 12ax³) + 27a⁴ (3x² - 2ax) + 27a⁶
= 27x⁶ - 108a³x³- 54ax⁵ + 36a²x⁴ + 81a²x⁴+ 36a⁴x² - 108a³x³ + 81a⁴x² - 54a⁵x + 27a⁶
= 27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³
+ 117a⁴x² - 54a⁵x + 27 a⁶

T_{2 =}T₁₊₁ =ⁿC₁ x^n-1. a=240 .....(i)
Similarly, T₃ =ⁿC₂ x^n-2. a² =720 ......(ii)
and T₄=ⁿC₃ x^n-3 a³ =1080......(iii)
On dividing. Eq. (ii) by Eq.(i), we get
\(\frac { { ^{ n }C }_{ 2 }.x^{ n-2 }.a^{ 2 } }{ { ^{ n }C }_{ 1 }.x^{ n-1 }.a } =\frac { 720 }{ 240 } \Rightarrow \frac { a }{ x } =\frac { 6 }{ n-1 } \)
On dividing Eq.(iii) by Eq. (ii) we get
\(\frac { a }{ x } =\frac { 9 }{ 2(n-2) } \Rightarrow Then,\quad \frac { 6 }{ n-1 } =\frac { 9 }{ 2(n-2) } \Rightarrow n=5\)
Now \(T_{ 2 }={ ^{ n }C }_{ 1 }x^{ 5-1 }.a=240\quad \Rightarrow 5.x^{ 4 }.a=240\)
\(\because \frac { a }{ x } =\frac { 6 }{ n-1 } =\frac { 6 }{ 5-1 } =\frac { 6 }{ 4 } =\frac { 3 }{ 2 }\)
\( \Rightarrow a=\frac { 3 }{ 2 } x\)
\(\therefore 5.x^{ 4 }.\frac { 3 }{ 2 } .x=240\)
Ans. x=2, a=3 and n=5

T₅ = ⁿC₄ x⁴, T₆ =  ⁿC₅ x⁵ and  T₇ = ⁿC₆ x⁶
\(\because \) Coefficient of these terms are in AP.
\(\therefore \) 2ⁿC₅ = ⁿC₄ + ⁿC₆
Solve it.

(i) (c)
(ii) (b)
(iii) (d) 
(iv) (b) 
(v) (d)