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

Click to See Answer :

Ans: (B) & (C)

Sol: $\large (4+3x)^{25}= 4^{25} (1+\frac{3}{4}x)^{25}$

Let (r + 1)^th term will have greatest coefficient,

$\large \frac{Coefficient \; of \; T_{r+1}}{Coefficient \; of \; T_r} \ge 1 $

$\large \frac{25_{C_r}(\frac{3}{4})^r}{25_{C_{r-1}}(\frac{3}{4})^{r-1}} \ge 1 $

$\large (\frac{25-r+1}{r}) \frac{3}{4} \ge 1 $

$\large r \le \frac{78}{7}$

Greatest possible value of r is 11

Coefficient of T₁₂ = 4²⁵ x ²⁵C₁₁ (3/4)¹¹

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

 

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.

Q: Which of the following is/are true in the expansion of (x + 2y + 3z)¹²

(A) The number of terms in the expansion is 91

(B) The sum of all the coefficient in the expansion is 126

(C) The number of terms in the expansion is 36

(D) The sum of all the coefficient in the expansion is 612

Sol. (x + 2y + 3z)¹²

= ¹²C₀ (x + 2y)¹² (3z)⁰ + ¹²C₁ (x + 2y)¹¹ (3z)¹ + ….+ ¹²C₁₂ (x + 2y)⁰ (3z)¹²

= 13 + 12 + 11 + …….1 = 91

Sum of coefficients = (1 + 2 + 3)¹² = 6¹²

Hence (A) and (D) are the correct answer.

Q: $\large \frac{1}{1!(n-1)!} – \frac{1}{3!(n-3)!} + \frac{1}{5!(n-5)!} – …is \; equal \; to $

(A) $\large \frac{2^{n-1}}{n!}$ for all n ∈ N

(B) $\large \frac{2^{n-1}}{n!}$ for odd values of n only

(C) $\large \frac{2^{n-1}}{n!}$ for even values of n only

(D) none of these

Sol. Given expression $\large = \frac{1}{n!}(n_{C_1} – n_{C_3} + n_{C_5} -…..) = \frac{2^{n/2}sin\frac{n\pi}{4}}{n!}$

Q: Given the integers r > 1, n > 2 , and coefficients of (3r)^th and (r+2)^nd term in the binomial expansion of (1+x)²ⁿ are equal, then

(A) n = 2r

(B) n =3r

(C) n = 2r+1

(D) None of these

Sol. Coefficients of (3r)^th and (r+2)^nd terms will be ²ⁿC_3r-1 and ²ⁿC_r+1

These are equal ⇒ (3r – 1) + (r + 1) = 2n ⇒ n = 2r

Hence (A) is the correct answer.

Q: The value of (n + 2) C₀ 2ⁿ⁺¹ – (n + 1) C₁2ⁿ + n. C₂ 2^n-1 – … is equal to

(A) 4 (1 + n)

(B) 4n

(C) 2n

(D) 2n + 4

Sol. (x – 1)ⁿ = C₀ xⁿ – C₁ x^n-1 + Cnx^n-2 – ….

x² (x – 1)n = C₀ xⁿ⁺² – C₁ xⁿ⁺¹ + C₂ xⁿ -….

differentiating with respect to x , we get

2x (x – 1)ⁿ + x²n (x – 1)^n-1 = (n + 2) C₀xⁿ⁺¹ – (n + 1) C₁ xⁿ + nC₂x^n-1 – ………

put, x = 2, we get

(n + 2) C₀ xⁿ⁺¹ – (n + 1) C₁2ⁿ + n. C₂2^n-1 – …. = 4 + 4n

Hence (A) is the correct answer.

Q: If the coefficients of x⁷ and x⁸ in $ (2 + \frac{x}{3})^n$ are equal, then n is

(A) 56

(B) 55

(C) 45

(D) 15

Sol. $\large n_{C_7} \frac{2^{n-7}}{3^7} = n_{C_8} \frac{2^{n-8}}{3^8} $

⇒ n = 55

Hence (B) is the correct answer.

Q: The term independent of x in $\large (\sqrt{\frac{x}{3}} + \sqrt{\frac{3}{2x^2}} )^{10} $ is

(A) 1

(B) 5/12

(C) ¹⁰C₁

(D) None of these

Sol. General term in the expansion is

$\large = 10_{C_r} (\frac{x}{3})^{r/2} (\frac{3}{2x^2})^{\frac{10-r}{2}}$

$\large = 10_{C_r} x^{\frac{3r}{2}-10} . \frac{3^{5-r}}{2^{(10-r)/2}}$

For constant term, 3r/2 = 10  ⇒ r = 20/3

which is not an integer. Therefore, there will be no constant term.

Hence (D) is the correct answer.