Binomial Theorem

Q: If ac > b² then the sum of the coefficients in the expansion of (a α²x² + 2b α x + c)ⁿ; (a, b, c, α ∈ R, n ∈ N ) is

(A) positive if a > 0

(B) positive if c > 0

(C) negative if a < 0, n is odd

(D) positive if c < 0, n is even

Sol. In the expansion of (a α²x² + 2b α x + c)ⁿ

the sum of the coefficients = (a α² + 2b α + c)ⁿ

Let f(α) = (a α² + 2b α + c)ⁿ

Its discriminant = 4b² – 4ac = 4(b² –ac ) < 0

Hence, f(α) < 0 or f(α) > 0 for all α ∈ R

If a > 0 then f(α) > 0 ⇒ (a α² + 2b α + c)ⁿ > 0

If c > 0 i.e. f(0) > 0 ⇒ f(α) > 0 ⇒ (a α² + 2b α + c)ⁿ > 0

If a < 0 then f(α) < 0 ⇒ (a α² + 2b α + c)ⁿ < 0 if n is odd

If c < 0 i.e. f(0) < 0 ⇒ f(α) < 0 ⇒ (a α² + 2b α + c)ⁿ > 0 if n is even.

Hence (A), (B), (C) and (D) are the correct answer.

Q: The largest coefficient in the expansion of (4+3x)²⁵ is

(A) ²⁵C₁₁.3²⁵ (4/3)¹¹

(B) ²⁵C₁₁.4²⁵ (3/4)¹¹

(C) ²⁵C₁₄.4¹⁴ (3)¹¹

(D) None of these

Q: The value of ⁿC₀ + ⁿ⁺¹C₁ + ⁿ⁺²C₂ + ….+ ^n+kC_k is equal to

(A) ^n+k+1C_k

(B) ^n+k+1C_n+1

(C) ^n+kC_n+1

(D) None of these

Sol. We have (1+x)ⁿ + (1+x)ⁿ⁺¹ + ….+ (1+x)^n+k

$\large = (1+x)^n \frac{(1+x)^{k+1}-1}{x} $

$\large = \frac{(1+x)^{n+k+1}-(1+x)^n}{x} $

Equating coefficients of xⁿ

ⁿC_n + ⁿ⁺¹C_n + ….+^n+kC_n = ^n+k+1C_n+1

⇒ ⁿC₀ + ⁿ⁺¹C₁ + ⁿ⁺²C₂ + ….+ ^n+kC_k

= ^n+k+1C_k = ^n+k+1C_n+1

Hence (A), (B) are the correct answer.

Q: If (1 + x)²ⁿ = a₀ + a₁x + a₂x² + …..+ a_2nx²ⁿ then

(A) a₀ + a₂ + a₄ + …. = (1/2)( a₀ + a₁ + a₂ + a₃ + ….)

(B) a_n + 1 < a_n

(C) a_n–3 = a_n+3

(D) none of these

Sol. a₀ + a₁ + a₂ + ….= 2²ⁿ

and a₀ + a₂ + a₄ + ….= 2^2n–1

a_n = ²ⁿC_n = the greatest coefficient, being the middle coefficient

a_n–3 = ²ⁿC_n–3 = ²ⁿC_{2n–(n –3)} = ²ⁿC_n+3 = a_n+3

Hence (A), (B) and ( C ) are the correct answer.