Coefficient of (2m + 1)th term = ¹⁰C_2m
and coefficient of (4m + 5)th term = ¹⁰C_{4m + 4}
Now, ¹⁰C_2m = ¹⁰C_{4m + 4}
\(\Rightarrow \frac { 10! }{ \left( 2m \right) !\left( 10-2m \right) ! } =\frac { 10! }{ \left( 4m+4 \right) !\left( 10-4m-4 \right) ! } \)
\(\Rightarrow \frac { 1! }{ \left( 2m \right) !\left( 10-2m \right) ! } =\frac { 1 }{ \left( 4m+4 \right) !\left( 6-4m \right) ! } \)
solve it.

(1 + x)³ (1 + x)⁶
= (1 + x³ + 3x + 3x²) (1 + ⁶C₁ x + ⁶C₂ x² - ⁶C₃​​​ ​​​​x³ + ⁶C₄​​​ ​​​​x⁴ - ⁶C₅​​​ ​​​​x⁵ + ⁶C6​​​ ​​​​x⁶)
Coefficient of x⁵ = - ⁶C₅​​​ + ⁶C₂  + 3 ⁶C₄​​​ - 3 ⁶C₃
= - 6 + 15 + 45 - 60 = - 6

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

n = 7, a = 2, x = 1

¹²C_r(2)^12-rx^24-3r

1157625

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 \(\frac { {-x y }^{ 9 } }{ 18 } \) .

-252

2 (x⁶+ 15x⁵ - 29x³ + 12x² + 3x - 1)

(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]

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 =​​​​​​​\(\frac{1}{3}\) 

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.