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

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.

## The largest coefficient in the expansion of (4+3x)^25 is

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

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.

## The value of nC0 + n+1C1 + n+2C2 + ….+ n+kCk is equal to

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

(A) n+k+1Ck

(B) n+k+1Cn+1

(C) n+kCn+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

nCn + n+1Cn + ….+n+kCn = n+k+1Cn+1

nC0 + n+1C1 + n+2C2 + ….+ n+kCk

= n+k+1Ck = n+k+1Cn+1

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

## If (1 + x)2n = a0 + a1x + a2x2 + …..+ a2nx2n then

Q: If (1 + x)2n = a0 + a1x + a2x2 + …..+ a2nx2n then

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

(B) an + 1 < an

(C) an–3 = an+3

(D) none of these

Sol. a0 + a1 + a2 + ….= 22n

and a0 + a2 + a4 + ….= 22n–1

an = 2nCn = the greatest coefficient, being the middle coefficient

an–3 = 2nCn–3 = 2nC2n–(n –3) = 2nCn+3 = an+3

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