Mathematical Induction and Binomial Theorem | JEE Advanced - Free MCQ Practice

Given that r and n are +ve integers such that r > 1, n > 2
Also in the expansion of  (1+ x)²ⁿ coeff. of  (3r)^th term = coeff. of (r + 2)^th term
⇒ ²ⁿC_3r–1= ²ⁿC_r+1
⇒ 3r –1 = r + 1 or 3r – 1+ r + 1 = 2n
⇒ r = 1 or 2r = n
But r  > 1      
∴       n = 2r

General term in the expansion is 

For coeff of x⁴, we should have

10 – 3r = 4  ⇒ r = 2

∴ Coeff  of  x⁴ =

The given expression is

We know by binomial theorem, that (x + a ) n + ( x – a ) ⁿ = 2 [ ⁿC₀ x ⁿ +  ⁿC ₂x ^{n – 2}a ² + ⁿC₄x^{n – 4}a₄ + ......]

∴ The given expression is equal to 2 [⁵C₀x⁵  +  ⁵C₂x₃(x³ – 1) +  ⁵C₄x (x³  – 1)²]
Max. power of x involved here is 7, also only +ve  integral powers of x are involved, therefore given expression is a polynomial of degree 7.

We have (1 + x)^m (1– x)ⁿ

= 1 + (m + n)x + 

Given,  m – n = 3 ....(1)

and

⇒ m² + n² - 2mn - (m +n) =-12

⇒ (m -n)² - (m +n) =-12
⇒ m +n = 9 + 12= 21 ....(2)
From (1) and (2), we get m = 12

NOTE THIS STEP : 

(a – b)ⁿ, n ≥ 5 In binomial expansion of above T5+ T6= 0
⇒ ⁿC₄ a^n–4 b₄ + ⁿC₅ a^n–5 b⁵ = 0

= Coeff of x^m in the expansion of product (1+ x)¹⁰ (1 + x)²⁰ =  Coeff of x^m  in the expansion of (1+ x)³⁰ = ³⁰C_m To get  max. value of given sum, 30Cm should be max. which is so when m = 30/2 = 15.

(1 + t²)¹²  (1 + t¹²) (1 + t²⁴) = (1 + t¹² + t²⁴ + t³⁶) (1 + t²)¹²
∴ Coeff. of t²⁴ = 1× Coeff. of  t²⁴ in (1+ t²)¹² + 1 × Coeff. of t¹² in (1 + t²)¹² + 1 × constant term in (1 + t²)¹² = ¹²C₁₂ + ¹²C₆ + ¹²C₀ = 1+¹²C₆ + 1=¹²C₆ + 2

Since  0 ≤ r ≤ n – 1

To find ³⁰C₀³⁰C₁₀ – ³⁰C₁³⁰ C₁₁ + ³⁰ C₂³⁰C₁₂ – .... + ³⁰C₂₀³⁰C₃₀
We know that (1 + x)³⁰ = ³⁰C₀ + ³⁰C₁x + ³⁰C₂x² + .... + ³⁰C₂₀x²⁰ + ....³⁰C₃₀x³⁰ ....(1)
(x – 1)³⁰ = ³⁰C₀x³⁰ –  ³⁰C₁x²⁹ +....+ ³⁰C₁₀x²⁰ – ³⁰C₁₁x¹⁹ + ³⁰C₁₂x¹⁸ +.... ³⁰C₃₀x⁰ ....(2)
Multiplying eqn (1) and (2), we get (x² – 1)³⁰ = (   ) × (   )
Equating the coefficients of x²⁰ on both sides,
we get ³⁰C₁₀ = ³⁰C₀³⁰ C₁₀ – ³⁰ C₁³⁰C₁₁  + ³⁰C₂³⁰C₁₂– ....
+ ³⁰C₂₀  ³⁰C₃₀
∴ Req. value is ³⁰C₁₀

Clearly  

Now expanding (1 + x )¹⁰ and (1 + x )²⁰ by bin om ial theorem and comparing the coefficients of x²⁰ in their product, on both sides, we get

 
= coeff of x²⁰ in (1 + x )³⁰ = ³⁰ C₂₀= ³⁰C₁₀

Again expending (1 + x )¹⁰ and ( x + 1)¹⁰ by bin omial theorem  and comparing the coefficients of x¹⁰ in their

product on both sides, we get

coeff of x¹⁰ in (1 + x) ²⁰ = ₂₀C₁₀

Substituting these values in equation (1), we get

Coeff. of x¹¹ in exp. of 

= (Coeff. of x^a ) × (Coeff. of x^b ) × (Coeff. of x^c )
Such that a + b + c = 11 Here a = 2m, b = 3n, c = 4p
∴ 2m + 3n + 4p = 11
Case I : m = 0, n = 1, p = 2
Case II : m = 1, n = 3, p = 0
Case III : m = 2, n = 1, p = 1
Case IV : m = 4, n = 1, p = 0
∴ Required coeff.

= 462 + 140 + 504 + 7 = 1113