Comparing \(\left( 2x+\frac { 3 }{ { x }^{ 2 } } \right) ^{ 9 }\)with (a+ b)ⁿ,
we have
a = 2x, b = \(\frac { 3 }{ { x }^{ 2 } } \) and n = 9
We know that
T_{r+ 1} = ⁿC_ra^n-r b^r
∴ T_r+1 = ⁹C_r (2x)^9-r\((\frac { 3 }{ { x }^{ 2 } } )^r\)
= ⁹C_r(2)^9-r (3)^rx^9-3r
Now 9 - 3r = 0 ⇒ 3r = 9 ⇒ r = 3
Thus T_{3+ 1} i.e. T₄ is the term independent of x, which is equal to ⁹C₃(2)⁶(3)³=19736

Here coefficient of x² in (1 + x)^m = ^mC₂=6
∴ \(\frac{m(m-1)}{2}=6\)
⇒ m²- m = 12
⇒ m²- m-12 = 0
⇒ m² - 4m + 3m - 12 = 0
⇒ m(m - 4) + 3(m - 4) = 0
⇒ (m - 4) (m + 3) = 0
Either m - 4 = 0 or m + 3 = 0
⇒ m = 4 or m =-3
So positive value of mis 4.

We know that coefficient of xⁿ in the expansion of (1 + x)²ⁿ
\(=^{ 2n }{ C }_{ n }=\frac { (2n)! }{ n!n! } =\frac { 2n(2n-1)! }{ n(n-1)!n! } \)
\(=2\frac { (2n-1)! }{ (n-1)!n! } ...(i)\)
Also the coefficient of xⁿ is the expansion of (1 + x)^2n-1
\(=^{ 2n-1 }{ C }_{ n }=\frac { (2n-1)! }{ (n-1)!n! } ...(ii)\)
From (t) and (ii) we see that coefficient of xⁿ in (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n - 1}.

The general term in the expansion of \(\left( 3x-\frac { 2 }{ x^{ 2 } } \right) ^{ 15 }\) is
\({ T }_{ r+1 }=^{ 15 }C_{ r }{ (3x) }^{ 15-r }\left( -\frac { 2 }{ x^{ 2 } } \right) ^{ r }=^{ 15 }C_{ r }{ (3x) }^{ 15-r }.x^{ 15-r }.(-1)^{ r }.\frac { { 2 }^{ r } }{ { x }^{ 2r } } \)
=¹⁵C_r(3)^15-r(-1)^r.2^r.x^15-3r
Putting r = 5 \({ T }_{ 6 }=^{ 15 }C_{ 5 }{ (3) }^{ 15-5 }\left( -1 \right) ^{ 5 }2^{ 5 }=-\frac { 15! }{ 5!10! } \times { 3 }^{ 10 }.{ 2 }^{ 5 }\)
\(=\frac { 15\times 14\times 13\times 12\times 11 }{ 5\times 4\times 3\times 2\times 1 } \times 59049\times 32=-5674372704\)

Here (1 + x² + 2x)²⁰ = [(1 + x)²]²⁰= (1 + x)⁴⁰
The general term in the expansion of (1 + x)⁴⁰ is
T_r+1 = 40C_rr^r
Putting r = 40 T₄₁ = ⁴⁰C₄₀x⁴⁰
∴ Co-efficient of x⁴⁰ = ⁴⁰C₄₀=1.

Comparing \(\left( x-\frac { { 1 } }{ 6^2 } \right) ^{ 24 }\)with(a+b)ⁿ,
we have 
a = x, b =\(\left( -\frac { 1 }{ { x }^{ 2 } } \right) \)and n=24
We know that 
T_r+1=ⁿC_ra^n-rb^r
\(\therefore \) T_r+1=²⁴C_rx^{24-r .\(\left( -\frac { 1 }{ { x }^{ 2 } } \right) ^r\)}
Now 24 -3r=6
\(\Rightarrow \)3r = 24 - 6 \(\Rightarrow \)  r = 6
\(\therefore \)  Coefficient of x⁶ = ²⁴C₆(- 1)⁶
= ²⁴C₆=​​​​​​​\(\frac { 24! }{ 18!6! } \)​​​​​​​

Comparing \(\left( 3x-\frac { { y }^{ 3 } }{ 6 } \right) ^{ 4 }\)with (a + b)ⁿ,
we have 
a = 3x,b = -\(\frac { { y }^{ 3 } }{ 6 } \)and n=4
Now   T₄=ⁿC₃(3x)^4-3\(\left( \frac { -{ y }^{ 3 } }{ 6 } \right) ^{ 3 }\)
  = ⁴C₃(3X)^{4- 3}\(\frac { { y }^{ 9 } }{ 216 } \)=\(\frac { {-x y }^{ 9 } }{ 18 } \)
 = Thus 4th term in the expansion of \(\left( 3x-\frac { { y }^{ 3 } }{ 6 } \right) ^{ 4 }\)is \(\frac { {-x y }^{ 9 } }{ 18 } \) .

Using binomial theorem for the expansion of (2x - 3)⁶ , we have
(2x -3)⁶ = ⁶C_o(2x)⁶ + ⁶C₁(2x)⁵(-3) + ⁶C₂(2x)⁴(-3)² + ⁶C₃(2x)³(-3)³ + ⁶C₄(2x)⁴(-3)⁴ + ⁶C₅2x⁵(-3)² + ⁶C₆(-3)⁶
= 64x⁶ + 6.32x⁵(-3) + 15.16x⁴.9 + 20.8x³(-27) + 15.4x².81 + 6.2x(-243) + 729
= 64x⁶ - 576x⁵ + 2160x⁴ - 4320x³ + 4860x² - 2916x + 729

(1+ 3X)⁶ (1 - X)⁵
= [⁶C₀ (3x)⁰ + ⁶C¹ (3X)¹ + ⁶C² (3X)²+⁶C₃ (3x)³ + ⁶C₄ (3X)⁴ + ⁶C₅ (3x)⁵+ ⁶C₆ (3x)⁶]
[⁵C₀ (-x)⁰ + ⁵C₁ (-x)¹+ ⁵C₂ (-x)² + ⁵C₃ (-x)³ + ⁵C₄ (-X)⁴+ ⁵C₅ (-X)⁵]
= [1 + 6 x 3x + 15 X 9x² + 20 x 27x³ + 15 x 81x⁴ + 6 x 243x⁵ + 729x⁶]
[1 - 5x + 10x² - 10x³ + 5x⁴- x⁵]
Coefficient of x⁵
= -1 + 90 -1350 + 5400 - 6075 + 1458 = 6948 -7426 = - 478

(x + 2y)⁸ + (x - 2y)⁸
= 2 [⁸C₀x⁸+ ⁸C₂x⁶ (2y)² + ⁸C₄ X⁴ (2y)⁴+ ⁸C₆x² (2y)⁶ + ⁸C₈(2y)⁸]
\(\therefore \) (x + a)ⁿ + (x - a)ⁿ
 2 [ⁿc₀ xⁿ + ⁿC₂ x^{n - 2} a²+ ^nC₄ x^{n - 4} a⁴ + ...]
\(\Rightarrow \)(x + 2y)⁸ + (x - 2y)B​​​​​​⁸
=[x⁸+ 28x⁶ X 4y² + 70 x⁴X 16y⁴ + 28x² x 64 y⁶ + 256y⁸]
= 2 [x⁸+ 112x⁶y² + 1120x⁴y⁴ + 1792x²y⁶ + 256y⁸]

Given equation of hyperbola is \(\frac { { y }^{ 2 } }{ 9 } -\frac { { x }^{ 2 } }{ 27 } =1\)
On comparing with \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } =1\) we get
a²=9, b²=27 or \(a=3,b=3\sqrt { 3 } \)
(ii) Centre (0,0)
(iii) Transverse axis, \(2a=2\times 3=6\)
(iv) Conjugate axis, \(2b=2\times 3\sqrt { 3 } =6\sqrt { 3 } \)
(v) Value of \(c=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =\sqrt { 9+27 } =\sqrt { 36 } =6\)
(vi) Vertices = \((0,\pm a)=(0,\pm 3)\)
(vii) Directions, \(y=\pm \frac { { a }^{ 2 } }{ c } =\pm \frac { 9 }{ 6 } =\pm \frac { 3 }{ 2 } \)
(viii) Foci = \((0,\pm c)=(0,\pm 6)\)
(ix) Eccentricity, \(e=\frac { c }{ a } =\frac { 6 }{ 3 } =2\)
(x) Length of latusrectum \(=\frac { { 2b }^{ 2 } }{ a } =\frac { 2\times 27 }{ 3 } =18\)

Here, value of index n is 10 (even), therefore there is only one middle term given by \({ T }_{ \frac { 10 }{ 2 } +1 }\) i.e. T_6.
Now, \({ T }_{ 6 }={ T }_{ 5+1 }=^{ 10 }C_{ 5 }\left( \frac { { 2x }^{ 2 } }{ 3 } \right) ^{ 10-5 }\left( \frac { 3 }{ { 2x }^{ 2 } } \right) ^{ 5 }[\because T_{ r+1 }=^{ n }{ C }_{ r }{ a }^{ n-r }{ b }^{ r }]\)
\(=^{ 10 }{ C }_{ 5 }\left( \frac { 2x^{ 2 } }{ 3 } \right) ^{ 5 }\left( \frac { 3 }{ 2x^{ 2 } } \right) ^{ 5 }=\frac { 10.9.8.7.6 }{ 5.4.3.2.1 } .\frac { { 2 }^{ 5 }.{ x }^{ 10 } }{ { 3 }^{ 5 } } .\frac { 3^{ 5 } }{ { 2 }^{ 5 }.{ x }^{ 10 } } =252\)
which is independent of x.

Given, T₃=1000
\(\Rightarrow ^{ 5 }{ C }_{ 2 }\left( \frac { 1 }{ x } \right) ^{ 5-2 }({ x }^{ \log { 10\quad x } })^{ 2 }=1000\)
\(\Rightarrow 10{ (x^{ \log { 10\quad x } })^{ 2 }\times { x }^{ -3 }=1000\Rightarrow x^{ 2log10\quad x }\times { x }^{ -3 }=100 }\)
\(\Rightarrow { x }^{ 2\quad log10\quad x-3 }={ 10 }^{ 2 }\)
\(\Rightarrow 2{ log }_{ 10 }\quad x-3={ log }_{ x }10^{ 2 }\)
\(\Rightarrow 2{ log }_{ 10 }\quad x-3=\frac { 2 }{ { log }_{ 10 }x } \)
\(\Rightarrow 2y-3=\frac { 2 }{ y } ,where\quad y=log_{ 10 }x\)
\( \Rightarrow 2y^{ 2 }-3y-2=0\Rightarrow (2y+1)(y-2)=0\)
\(\Rightarrow y=2\quad or\quad y=-\frac { 1 }{ 2 } \Rightarrow { log }_{ 10 }x=2\quad or\quad { log }_{ 10 }x=-\frac { 1 }{ 2 } \)
\( \Rightarrow x={ 10 }^{ 2 } \ or \ x=10^{ -1/2 }\)
\(\therefore \) x = 100 or \(x=\frac { 1 }{ \sqrt { 10 } } \)

The 5th term from the end in the expansion of \(\left( x-\frac { 1 }{ { x }^{ 2 } } \right) ^{ 12 }\) = The (12-5+2)^th 
i.e. 9th term from the beginning in the expansion of \(\left( x-\frac { 1 }{ { x }^{ 2 } } \right) ^{ 12 }\)
[\(\because \) pth term from the end = (n-p+2)th term from the beginning]
Now, \({ T }_{ 8+1 }=^{ 12 }{ C }_{ 8 }(x)^{ 12-8 }\left( -\frac { 1 }{ x^{ 2 } } \right) ^{ 8 }\)
\(=^{ 12 }{ C }_{ 4 }{ x }^{ 4 }(-1)^{ 8 }.\frac { 1 }{ x^{ 16 } } \)
\(=\frac { 12\times 11\times 10\times 9 }{ 4\times 3\times 2\times 1 } \times x^{ 4-16 }=495{ x }^{ -12 }=\frac { 495 }{ { x }^{ 12 } } \)

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