T_r+1=¹⁰C_r (-1)^r x^2r
For coefficient of x¹⁰, put 2r=10 \(\Rightarrow \)r=5, then
Coefficient of x¹⁰=¹⁰C₅ (-1)⁵ and T^'_r+1=¹⁰C_r x^10-2r (-2r)^r
For independent of x, put 10-2r=0, \(\Rightarrow \) r=5,
Then, T₆' =¹⁰C₅ (-2)⁵
Now  \(\frac { { ^{ 10 }C }_{ 5 }(-1)^{ 5 } }{ { ^{ 10 }C }_{ 5 }(-2)^{ 5 } } =\frac { 1 }{ 32 } \)

\(LHS=(1+y+{ y }^{ 2 }+...)^{ 2 }=(\frac { 1 }{ 1-y } )^{ 2 }=\left( 1-y \right) ^{ -2 }=1+2y+3{ y }^{ 2 }+...=RHS\)

General term in \({ \left( { 2x }^{ 2 }-\frac { 3 }{ x } \right) }^{ 11 }\), 
T_r+1 = ¹¹C_r (2x²)^11-r \({ \left( \frac { 3 }{ x } \right) }^{ r }\)
= ¹¹C_r 2^11-rx^22-2r-r (-3)^r
for x⁸, 22 - 2r - r = 8 \(\Rightarrow \) 3r = 22 - 8 \(\Rightarrow \) r = \(\frac { 14 }{ 3 } \)  
It is not an integer.
So, no term involving x⁸ exist.

General term in\({ \left( { ax }^{ 2 }+\frac { 1 }{ bx } \right) }^{ 11 }\) ,
T_r+1 = ¹¹C_r a^11-r b^-r x^22-3x
for x⁷, 22 - 3r = 7 \(\Rightarrow \) r = 5
\(\therefore \)  Coefficient of x⁷ in \({ \left( { ax }^{ 2 }+\frac { 1 }{ bx } \right) }^{ 11 }\)  = ¹¹C₅ a⁶ b^-5
General term in\({ \left( { ax }-\frac { 1 }{ b{ x }^{ 2 } } \right) }^{ 11 }\) ,
T_r+1 = ¹¹C_r a^11-r (-1)^r b^-r x^11-3x
for x⁷, 11 - 3r = -7 \(\Rightarrow \) r = 6
\(\therefore \) Coefficient of x^-7 in \({ \left( { ax }-\frac { 1 }{ b{ x }^{ 2 } } \right) }^{ 11 }\)  =  ¹¹C₆ a⁵ b^-6(-1)⁶
According to the question, 
¹¹C₅ a⁶ b^-5 = ¹¹C₆ a⁵ b^-6(-1)⁶
\(\Rightarrow \) ¹¹C₅ ab = ¹¹C₅ \(\Rightarrow \) ab = 1

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
\(\Rightarrow 6{ \alpha }^{ 2 }+20\alpha ^{ 3 }=0\Rightarrow 2{ \alpha }^{ 2 }\left( 3+10\alpha \right) =0\)
\( \Rightarrow \alpha =0,\frac { -3 }{ 10 } \)

We have (1 + x)²ⁿ
∴T_m+1=²ⁿC_m(x)^m
∴ Coefficient of (m + 1)^th term = ²ⁿC_m
Also T_m+3 = ²ⁿC_m+2 (x)^{m + 2}
∴ Coefficient of (m + 3)^th term = ²ⁿC_{m + 2}
It is given that
²ⁿC_m =²ⁿC_m+2
∴ m + (m + 2) = 2n
⇒ 2m + 2 = 2n
⇒ m+ 1= n.

Let r^th, (r + 1)^th and (r + 2)^th be three consecutive terms in the expansion of (1 + x)ⁿ. Then their coefficients are ⁿC _r-1,ⁿ C _r and ⁿC_r+1 respectively.
∴ ⁿC_r-1:ⁿC_r:ⁿC_r+1=1:6:30
Now \(\frac { ^{ n }{ C }_{ r-1 } }{ ^{ n }{ C }_{ r } } =\frac { 1 }{ 6 } \)
\(\Rightarrow \frac { r }{ n-r+1 } =\frac { 1 }{ 6 } \)
⇒ n -7r =-1 ....(i)
Also \(\frac { ^{ n }{ C }_{ r } }{ ^{ n }{ C }_{ r+1 } } =\frac { 6 }{ 30 } \)
\(\Rightarrow \frac { r+1 }{ n-r } =\frac { 1 }{ 5 } \)
n - 6r = 5 ...(ii)
Solving (i) and (ii), we have
n = 41 and r = 6.

It is given that T₄ = \(\frac{20}{27}\) 
Comparing\(\left( ax+\frac { 1 }{ x } \right) ^{ n }\) with (A+ B)ⁿ, we have
A=ax and B=\(\frac{1}{x}\)
∴ T₄ = ⁿC₃ A^n-3\((\frac{1}{x})^3\)
= ⁿC₃ (ax)^n-3\((\frac{1}{x})^3\)
∴ ⁿc₃ (ax)^n-3\((\frac{1}{x})^3=\frac{20}{27}\)
∴ = ⁿc₃ a^n-3 x^n-6=\(\frac{20}{27}...(i)\)
Now n - 6 = 0 ⇒ n = 6
Putting n = 6 in (i), we have
⁶C₃ a^{6- 3}x^{6 - 6}=​​​​​​​\(\frac{20}{27}\)
⇒ 20a³=\(\frac{20}{27}\)
⇒ a³=​​​​​​​\(\frac{1}{27}\)
⇒ a=\(\frac{1}{3}\)
Thus n = 6 and a =​​​​​​​\(\frac{1}{3}\) 

Here (x + c)ⁿ
=ⁿC₀ xⁿ+ⁿc₁ x^n-1a+ⁿC₂ x^n-2a² +... +ⁿC_naⁿ 
= P + Q....(i)
where P= ⁿC₀ xⁿ + ⁿC₂ x^n-2a² + ...
Q=ⁿC₁ x^{n - 1}a + ⁿC₃ x^{n- 3} a³ + ...
Also (x-a)^n
nC₀ xⁿ - ⁿC₁ x^{n - 1}a + ⁿC₂ x^n-2a² +...+(-1)^{n n}C_naⁿ 
=P - Q....(ii)
(i) Squaring and adding (i) and (ii), we have
(x + a)²ⁿ + (x - a)²ⁿ = (P + Q)² + (P - Q)²
= P² + Q² + 2PQ + p² +Q²-2PQ
= 2P²+ 2Q² =2 (P²+ Q²)
(ii)Multiplying (i) and (ii), we have
(x + a)ⁿ (x - a)ⁿ = (P + Q) (P - Q)
(x²-a²)ⁿ = P²- Q².

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