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.

We have
T₁= ⁿC₀ aⁿb⁰=729 ....... (i)
T₂ = ⁿC₁ a^n-1 b = 7290 ..........(ii)
T₃ = ⁿC₂ a^n-2 b² = 30375 .........(iii)
From (i) an = 729 ........(iv)
From (ii) n^a-1+b= 7290 .......(v)
From (iii) \(\frac { n(n-1) }{ 2 } { a }^{ n-2 }{ b }^{ 2 }=30375.....(vi)\)
Multiplying (iv) and (vi), we get
\(\frac { n(n-1) }{ 2 } { a }^{2 n-2 }{ b }^{ 2 }=729\times 30375...(vii)\)
Squaring both sides of (v), we get
n²a^2n-2b² = (7290)² ...(viii)
Dividing (vii) by (viii), we get
\(\frac { n(n-1){ a }^{ 2n-2 }{ b }^{ 2 } }{ { { 2n }^{ 2 }a }^{ 2n-2 }{ b }^{ 2 } } =\frac { 729\times 30375 }{ 7290\times 7290 } \)
\(\Rightarrow \frac { (n-1) }{ 2n } =\frac { 30375 }{ 72900 } \Rightarrow \frac { n-1 }{ 2n } =\frac { 5 }{ 12 } \)
⇒12n - 12 = 10n
⇒2n = 12 ⇒ n = 6
From (iv) a⁶ = 729⇒a⁶ = (3)⁶ ⇒ a = 3
From (v) 6 \(\times\) 35 \(\times\)b = 7290 ⇒ b = 5
Thus a = 3, b = 5 and n = 6.

Here n = 7, which is odd.
So the middle terms are \(\left( \frac { 7+1 }{ 2 } \right) th,\left( \frac { 7+1 }{ 2 } +1 \right) th\) are 4^th and 5^th terms.
The general term in the expansion of\(\left( 3-\frac { { x }^{ 3 } }{ 6 } \right) ^{ 7 }\) is
\({ T }_{ r+1 }=^{ 7 }{ C }_{ r }{ (3) }^{ 7-r }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ r }....(i)\)
Putting r = 3 and 4 in (i)
\(\therefore \quad { T }_{ 4 }=^{ 7 }C_{ 3 }{ (3) }^{ 7-3 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 3 }{ (3) }^{ 4 }.{ (-1) }^{ 3 }.\frac { { x }^{ 9 } }{ { (6) }^{ 3 } } \)
\(=35\times 81\times -\frac { { x }^{ 9 } }{ 216 } =-\frac { 105 }{ 8 } { x }^{ 9 }\)
Now \({ T }_{ 5 }=^{ 7 }C_{ 4 }{ (3) }^{ 7-4 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 4 }{ (3) }^{ 3 }.(-1)^{ 4 }\frac { { x }^{ 12 } }{ { (6) }^{ 4 } } \)
\(=35\times 27\times \frac { { x }^{ 12 } }{ 1296 } =\frac { 35 }{ 48 } { x }^{ 12 }\) 

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

We have \(\left[ 1+\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \right] ^{ 4 }\)
=\(\left[ 1+\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \right] ^{ 4 }\)
=⁴C₀+⁴C₁\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \)+⁴C₂\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^2\)+⁴C₃\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^3\)+⁴C₄\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^4\)
=1+4\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \)+6\(\left( \frac { x^{ 2 } }{ 2 } -\frac { 2 }{ { x }^{ 2 } } -2 \right) \)+4\(\left( \frac { { x }^{ 3 } }{ 8 } -\frac { 8 }{ { x }^{ 3 } } -\frac { 3x }{ 2 } +\frac { 6 }{ x } \right) +\left[ ^{ 4 }C_{ 0 }\left( \frac { x }{ 2 } \right) ^{ 4 }-^{ 4 }C_{ 1 }\left( \frac { x }{ 2 } \right) ^{ 3 }\left( \frac { 2 }{ x } \right) +^{ 4 }C_{ 2 }\left( \frac { x }{ 2 } \right) ^{ 2 }\left( \frac { 2 }{ x } \right) ^{ 2 }-^{ 4 }C_{ 3 }\left( \frac { x }{ 2 } \right) \left( \frac { 2 }{ x } \right) ^{ 3 }+^{ 4 }C_{ 4 }\left( \frac { x }{ 2 } \right) ^{ 4 } \right] \)
=1+\(\left( 2x-\frac { 8 }{ x } \right) +\left( \frac { 3 }{ 2 } { x }^{ 2 }+\frac { 24 }{ { x }^{ 2 } } -12 \right) +\left( \frac { { x }^{ 3 } }{ 2 } -\frac { 32 }{ { x }^{ 3 } } -6x+\frac { 24 }{ x } \right) +\left( \frac { { x }^{ 4 } }{ 16 } -{ x }^{ 2 }+6\frac { 16 }{ { x }^{ 2 } } +\frac { 16 }{ { x }^{ 4 } } \right) \)
=-5-4x + \(\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 2 } +\frac { { x }^{ 4 } }{ 16 } +\frac { 16 }{ x } +\frac { 8 }{ { x }^{ 2 } } +\frac { 32 }{ { x }^{ 3 } } -\frac { 16 }{ { 4 }^{ 4 } } \)

Putting a²=x and \(\sqrt { { a }^{ 2 }-1 } \) =y, we have
\(\left( { a }^{ 2 }+\sqrt { { a }^{ 2 }-1 } \right) ^{ 4 }+({ a }^{ 2 }-\sqrt { { a }^{ 2 }-1)^{ 4 } } \)
=(x+y)⁴+(x-y)⁴
=[⁴C₀x⁴+⁴C₁x³y+⁴C₂x²y²+⁴C₃xy³+⁴C₄y⁴]+[⁴C₀x^4-4C₁x³y+⁴C₂x²y^2-4C₃xy^3-4C₄y⁴]
=2[⁴C₀x⁴+⁴C₂x²y²+⁴C₄y⁴]
=2[x⁴+6x²y²+y⁴]
=2[a²)⁴+6(a²)²\(\left( \sqrt { { a }^{ 2 }-1 } \right) ^{ 2 }\)+\(\left( \sqrt { { a }^{ 2 }-1 } \right) ^{ 4 }\)]
=2[a⁸+6a⁴(a²-1)+(a²-1)²]
=2[a⁸+6a⁶-6a⁴+a⁴-2a²+1]
=2[a⁸+6a⁶-5a⁴-2a²+1]