If the constant term in the expansion of $\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}$ is $2^{\mathrm{k}} . l$, where $l$ is an odd integer, then the value of k is equal to :

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Binomial Theorem

If the constant term in the expansion of

(3 x^{3} - 2 x^{2} + \frac{5}{x ^{5}})^{10}

2^{k} . l

, where

l

is an odd integer, then the value of k is equal to :

Answer: (D)

The general term in the multinomial expansion is:

T_{r + 1} = \frac{10 !}{r _{1} ! r _{2} ! r _{3} !} (3)^{r_{1}} (- 2)^{r_{2}} (5)^{r_{3}} (x)^{3 r_{1} + 2 r_{2} - 5 r_{3}}

For the constant term:

3 r_{1} + 2 r_{2} - 5 r_{3} = 0

... (1)

Also:

r_{1} + r_{2} + r_{3} = 10

... (2)

From (1) and (2):

3 r_{1} + 2 (10 - r_{3}) - 5 r_{3} = 0

3 r_{1} + 20 = 7 r_{3}

Trying integer solutions:

(r_{1}, r_{2}, r_{3}) = (1, 6, 3)

Constant term

= \frac{10 !}{1 ! \cdot 6 ! \cdot 3 !} (3)^{1} (- 2)^{6} (5)^{3}

= \frac{10 \cdot 9 \cdot 8 \cdot 7}{6} \cdot 3 \cdot 64 \cdot 125 = 2^{9} \cdot 3^{2} \cdot 5^{4} \cdot 7

Therefore,

k = 9

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option