The general term of (1+x)²ⁿ is T_r+1 =²ⁿC_rx^r
Then, T₂=²ⁿC₁x¹ , T₃=²ⁿC₂x² and T₄ =²ⁿC₃x³
Given, coefficients of T₂ , T₃ and T_{4 are in AP.}
i.e.²ⁿC₁ ,²ⁿC₂ and ²ⁿC₃ are AP.
\(\therefore 2\times { ^{ 2n }C }_{ 2 }={ ^{ 2n }C }_{ 1 }+{ ^{ 2n }C }_{ 3 } \Rightarrow 2=\frac { { ^{ 2n }C }_{ 1 } }{ { ^{ 2n }C }_{ 2 } } +\frac { { ^{ 2n }C }_{ 3 } }{ { ^{ 2n }C }_{ 2 } } \)
\(\Rightarrow 2=\frac { (2n)! }{ 1!(2n-1)! } \times \frac { 2!(2n-2)! }{ (2n)! } +\frac { (2n)! }{ 3!(2n-3)! } \times \frac { 2!(2n-2)! }{ (2n)! }\)
\( \Rightarrow 2=\frac { 2(2n-2)! }{ (2n-1)(2n-2)! } +\frac { 2(2n-2)(2n-3)! }{ 6(2n-3)! } \)
\(\Rightarrow 2=\frac { 2 }{ 2n-1 } +\frac { 2n-2 }{ 3 } \Rightarrow 2=\frac { 6+(2n-2)(2n-1) }{ 3(2n-1) } \)
\(\Rightarrow 6(2n-1)=6+4n^{ 2 }-2n-4n+2\)
\(\Rightarrow 12n-6=8+4n^{ 2 }-6n\)
\( \Rightarrow 4n^{ 2 }-6n-12n=-6-8 \Rightarrow 4n^{ 2 }-18n=-14\)
\(\therefore \quad 2n^{ 2 }-9n=-7\)  [dividing by 2]

Given series is (1+x)¹⁰⁰⁰ + x(1+x)⁹⁹⁹ +x² (1+x)⁹⁹⁸ +...+x¹⁰⁰⁰
Here first term, a=(1+x)¹⁰⁰⁰
and common ratio, \(r=\frac { x(1+x)^{ 99 } }{ (1+x)^{ 1000 } } \)
\(=\frac { x^{ 2 }(1+x)^{ 998 } }{ x(1+x)^{ 999 } } =\frac { x }{ 1+x } \)
Thus, the given series is a geometric series with the common ratio \(\frac { x }{ 1+x } \)
Now, sum of geometric series = \(\frac { a(1-r^{ n }) }{ 1-r } \)
\(\frac { (1+x)^{ 1000 }\left[ 1-\left( \frac { x }{ 1+x } \right) ^{ 1001 } \right] }{ \left[ 1-\left( \frac { x }{ 1+x } \right) \right] } =\frac { (1+x)^{ 1000 }-\frac { x^{ 1001 } }{ 1+x } }{ \frac { 1+x-x }{ 1+x } } \)
\(=(1+x)^{ 1001 }-x^{ 1001 }\)
Hence, coefficient of x⁵⁰ in {(1+x)¹⁰⁰¹-x¹⁰⁰¹} is given by 
¹⁰⁰¹C₅₀ \(=\frac { 1001! }{ 50!951! } \)

We have , foci of the hyperbola lies on X-axis. So, the equation of hyperbola is
    \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) ---- (i)
Foci \(\equiv \left( \pm c,0 \right) \equiv \left( \pm 2,0 \right) \Rightarrow c=2\)
Eccentricity of the hyperbola, e=3/2
We know that, c=ae
\(\therefore \quad 2=a\left( \frac { 3 }{ 2 } \right) \Rightarrow a=\frac { 4 }{ 3 } \\ Also\quad \quad { c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\\ \Rightarrow \quad ({ 2 })^{ 2 }=\left( \frac { 4 }{ 3 } \right) ^{ 2 }+{ b }^{ 2 }\Rightarrow { b }^{ 2 }=\frac { 20 }{ 9 } \\ Thus,\quad { a }^{ 2 }=\frac { 16 }{ 9 } ,{ b }^{ 2 }=\frac { 20 }{ 9 } \)
Hence, equation of hyperbola is
\(\frac { 9{ x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 20 } =1\quad \quad \quad [from\quad Eq.(i)]\\ \Rightarrow \quad 45{ x }^{ 2 }-36{ y }^{ 2 }=80\)

The general term in the expansion of \(\left( x^{ 2 }+\frac { 1 }{ x } \right) ^{ 2n }\) is
\(T_{ r+1 }=^{ 2n }C_{ r }(x^{ 2 })^{ 2n-r }\left( \frac { 1 }{ x } \right) ^{ r }\)
\(=^{ 2n }{ C }_{ r }\quad x^{ 4n-2r }\quad \frac { 1 }{ { x }^{ r } } \)
\(\Rightarrow T_{ r+1 }=^{ 2n }{ C }_{ r }\quad x^{ 4n-3r }\)
On putting 4n-3r = p, we get
3r = 4n-p \(\Rightarrow r=\frac { 4n-p }{ 3 } \)
Now, coefficient of x^p
\(=^{ 2n }C\left( \frac { 4n-p }{ 3 } \right) =\frac { (2n)! }{ \left( \frac { 4n-p }{ 3 } \right) !\left( 2n-\frac { 4n-p }{ 3 } \right) ! } \)
\(=\frac { (2n)! }{ \left( \frac { 2n+p }{ 3 } \right) !\left( \frac { 4n-p }{ 3 } \right) ! } \)
Hence proved.

Here, n=10, which is an even number.
\(\therefore \) Middle term is given by \(\left( \frac { 10 }{ 2 } +1 \right) \)^th term, i.e. 6th term.
Let \({ T }_{ r+1 }\) be the general term.
Then, \({ T }_{ r+1 }=^{ 10 }{ C }_{ r }\left( \frac { 1 }{ x } \right) ^{ 10-r }(x\sin { x } )^{ r }\)
\(=^{ 10 }{ C }_{ r }{ x }^{ r-10+r }sin^{ r }x=^{ 10 }{ C }_{ r }{ x }^{ 2r-10 }{ sin }^{ r }x\)
Now, \({ T }_{ 6 }={ T }_{ 5+1 }=^{ 10 }{ C }_{ 5 }x^{ 10-10 }sin^{ 5 }x=^{ 10 }{ C }_{ 5 }sin^{ 5 }x\)
\(\because \) Middle term=\(7\frac { 7 }{ 8 } \)
\(\therefore \) \(^{ 10 }{ C }_{ 5 }sin^{ 5 }x=\frac { 63 }{ 8 } \Rightarrow { sin }^{ 5 }x=\frac { 63 }{ 8\times 252 } =\frac { 1 }{ 8\times 4 } =\frac { 1 }{ 32 } \)
\(\Rightarrow sinx=\frac { 1 }{ 2 } \Rightarrow sinx=sin\frac { \pi }{ 6 }\)
\(\therefore x=n\pi +(-1)^{ n }\frac { \pi }{ 6 } \)

We have,\({ 25 }^{ 15 }=\left( 26-1 \right) ^{ 15 }\)
\(=^{ 15 }{ C }_{ 0 }\left( 26 \right) ^{ 15 }-^{ 15 }{ C }_{ 1 }\left( 26 \right) ^{ 14 }\left( 1 \right) ^{ 1 }+^{ 15 }{ C }_{ 2 }\left( 26 \right) ^{ 13 }\left( 1 \right) ^{ 2 }-.....-^{ 15 }{ C }_{ 15 }\)
\( =^{ 15 }{ C }_{ 0 }\left( 26 \right) ^{ 15 }-^{ 15 }{ C }_{ 1 }\left( 26 \right) ^{ 14 }+....-1-12+12\)
\(=\left( ^{ 15 }{ C }_{ 6 }2\times 13\times \left( 26 \right) ^{ 14 }-^{ 15 }{ C }_{ 1 }2\times 13\times \left( 26 \right) ^{ 13 }+.....13 \right) +12\)
\(=13\left( ^{ 15 }{ C }_{ 0 }\times 2\times \left( 26 \right) ^{ 14 }-^{ 15 }{ C }_{ 1 }\times 2\times \left( 26 \right) ^{ 13 }+......-1 \right) +12\)

\((1+x+{ x }^{ 2 }+{ x }^{ 3 })^{ 11 }=[(1+x)+{ x }^{ 2 }(1+x)]^{ 11 }\)
\(=(1+x)^{ 11 }.(1+x^{ 2 })^{ 11 }\)
\(=(1+11x+{ 55x }^{ 2 }+{ 165x }^{ 3 }+{ 330x }^{ 4 }+...)\quad (1+{ 11x }^{ 2 }+{ 55 }x^{ 4 }+....)\)
\(\therefore \) Coefficient of x⁴ = 55+605+330=990
Ans: 990

Let E = \(\left( { x }^{ 2 }-\sqrt { 1-{ x }^{ 2 } } \right) ^{ 4 }+\left( { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \right) ^{ 4 }\)
Put \(\sqrt { 1-{ x }^{ 2 } } =y,\quad \) , we get
\(E=\left( { x }^{ 2 }-y \right) ^{ 4 }+\left( { x }^{ 2 }+y \right) ^{ 4 }\)
\(=2[^{ 4 }{ C }_{ o }\left( { x }^{ 2 } \right) ^{ 4 }{ y }^{ o }+^{ 4 }{ C }_{ 2 }\left( { x }^{ 2 } \right) ^{ 2 }{ y }^{ 2 }+^{ 4 }{ C }_{ 4 }\left( { x }^{ 2 } \right) ^{ o }{ y }^{ 4 }]\)
\(\left[ if\quad n\quad is \quad even,then(x-a)^{ n }+(x+a)^{ n }=2\left\{ ^{ n }{ C }_{ o }{ x }^{ n }{ a }^{ o }+^{ n }{ C }_{ 2 }{ x }^{ n-2 }{ a }^{ 2 }+^{ n }{ C }_{ 4 }{ x }^{ n-4 }{ a }^{ 4 }+... \right\} \right] \)
\(=2[1\times { x }^{ 8 }\times 1+6\times { x }^{ 4 }{ y }^{ 2 }+1\times 1\times { y }^{ 4 }]\)
\(=2[{ x }^{ 8 }+6{ x }^{ 4 }(1-{ x }^{ 2 })+(1-{ x }^{ 2 })^{ 2 }]\quad [put\quad y=\sqrt { 1-{ x }^{ 2 } } \)
\(=2({ x }^{ 8 }+6{ x }^{ 4 }-6{ x }^{ 6 }+1+{ x }^{ 4 }-2{ x }^{ 2 })\)
\(=2{ x }^{ 8 }-12{ x }^{ 6 }+14{ x }^{ 4 }-4{ x }^{ 2 }+2\)

\(Coefficient\quad of\quad { x }^{ n }\quad in\quad the\quad expansion\quad of\quad { \left( 1+x \right) }^{ 2n }=^{ 2n }{ C_{ n } }\)
\(Coefficient\quad of\quad { x }^{ n }\quad in\quad the\quad expansion\quad of\quad { \left( 1+x \right) }^{ 2n-1 }=^{ 2n-1 }{ C_{ n-1 } }\)
\( Required\quad ratio=\frac { ^{ 2n }{ C_{ n } } }{ ^{ 2n-1 }{ C_{ n-1 } } } =\frac { \left( 2n \right) ! }{ n!n! } \times \frac { \left( n-1 \right) !n! }{ \left( 2n-1 \right) ! } =\frac { 2 }{ 1 } or\quad 2:1\)

\({ T }_{ 7 }=^{ n }{ { C }_{ 6 } }{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ n-6 }{ \left( \frac { 1 }{ \sqrt [ 3 ]{ 3 } } \right) }^{ 6 }=^{ n }{ { C }_{ 6 } }{ 2 }^{ \frac { n-6 }{ 3 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }\) 
and 7th term from the end = (n + 1) - (7 - 1)th term from beginning =(n - 5)th term from beginning.
\(=^{ n }{ { C }_{ n-6 } }{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ n-n+6 }{ \left( \frac { 1 }{ \sqrt [ 3 ]{ 3 } } \right) }=^{ n }{ { C }_{ n-6 } }{ 2 }^{ 2 }{ \left( \frac { 1 }{ 3 } \right) }^{ \frac { n-6 }{ 3 } }\) 
Now, \(\frac { ^{ n }{ { C }_{ 6 } }{ 2 }^{ \frac { n-6 }{ 3 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 2 } }{ ^{ n }{ { C }_{ n-6 } }{ 2 }^{ 2 }{ \left( \frac { 1 }{ 3 } \right) } } =\frac { 1 }{ 6 } \Rightarrow \frac { N-12 }{ 3 } =-1\Rightarrow N=9\)