JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Binomial Theorem

Let

α

be the constant term in the binomial expansion of

(x - \frac{6}{x ^{3/2}})^{n}

n \leq 15

. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of

x^{- n}

λ α

, then

λ

is equal to ______.

Answer: 36

Solution

The general term is

T_{k + 1} = (k n) (x)^{n - k} (- \frac{6}{x ^{3/2}})^{k} = (k n) (- 6)^{k} x^{(n - k) /2 - 3 k /2} = (k n) (- 6)^{k} x^{(n - 4 k) /2}

. For the constant term,

n - 4 k = 0

, so

k = n /4

. The constant term is

α = (n /4 n) (- 6)^{n /4}

. The sum of all coefficients is

(1 - 6)^{n} = (- 5)^{n}

. The sum of remaining terms is

(- 5)^{n} - α = 649

. By observation,

(- 5)^{4} = 625

, so

625 + 24 = 649

, giving

n = 4

and

α = (1 4) (- 6)^{1} = - 24

. The coefficient of

x^{- n} = x^{- 4}

occurs when

(n - 4 k) /2 = - 4

, i.e.,

4 - 4 k = - 8

, so

k = 3

. This coefficient is

(3 4) (- 6)^{3} = 4 (- 216) = - 864 = λ (- 24)

, giving

λ = 36

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer