The number of positive integers k such that the constant term in the binomial expansion of $\left(2x^{3}+\frac{3}{x^{k}}\right)^{12}$, $x \neq 0$ is $2^{8} \cdot \ell$, where $\ell$ is an odd integer, is ____.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Binomial Theorem

The number of positive integers k such that the constant term in the binomial expansion of

(2 x^{3} + \frac{3}{x ^{k}})^{12}

x \neq = 0

2^{8} \cdot ℓ

, where

ℓ

is an odd integer, is ____.

Answer: 2

The general term of

(2 x^{3} + \frac{3}{x ^{k}})^{12}

(r 12) 2^{r} 3^{12 - r} x^{3 r - k (12 - r)}

, so a constant term needs

3 r = k (12 - r)

, i.e.

k = \frac{3 r}{12 - r}

. This gives positive integers

k

for

r = 3, 6, 8, 9, 10, 11

(with

k = 1, 3, 6, 9, 15, 33

). The constant term for a given

r

(r 12) 2^{r} 3^{12 - r}

, and since

3^{12 - r}

is odd its power of two is

v_{2} ((r 12)) + r

. For the term to equal

2^{8} \cdot ℓ

with

ℓ

odd this must equal exactly

8

. Here

r = 6

gives

v_{2} (6 12) + 6 = 2 + 6 = 8

and

r = 8

gives

v_{2} (8 12) + 8 = 0 + 8 = 8

, while the remaining cases give

5, 11, 11, 13

. Only

r = 6

and

r = 8

qualify, i.e.

k = 3

and

k = 6

, so exactly

2

positive integers

k

work.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer