Binomial Theorem

Binomial Theorem

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

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

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.

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

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.