JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The constant term in the expansion of $(2 x + \frac{1}{x ^{7}} + 3 x^{2})^{5}$ is

Choose the correct answer:

Explanation

\left( 2x + \frac{1}{x^7} + 3x^2 \right)^5

The general term is given by:

T = \frac{5!}{n_1! n_2! n_3!} (2x)^{n_1} \left( \frac{1}{x^7} \right)^{n_2} (3x^2)^{n_3}

where $n_1 + n_2 + n_3 = 5$ and $n_1, n_2, n_3 \ge 0$ .

Simplifying the term:

T = \frac{5!}{n_1! n_2! n_3!} \cdot 2^{n_1} \cdot 3^{n_3} \cdot x^{n_1 - 7n_2 + 2n_3}

For the constant term, the exponent of $x$ must be zero:

n_1 - 7n_2 + 2n_3 = 0 \quad \dots (1)

n_1 + n_2 + n_3 = 5 \quad \dots (2)

Subtracting (1) from (2):

(n_1 + n_2 + n_3) - (n_1 - 7n_2 + 2n_3) = 5 - 0

8n_2 - n_3 = 5

n_3 = 8n_2 - 5

Possible values for $n_2$ :

If $n_2 = 1$ : $n_3 = 8(1) - 5 = 3$
Then $n_1 = 5 - 1 - 3 = 1$

(Other values of $n_2$ lead to $n_i < 0$ or $n_i > 5$ )

The only solution is $(n_1, n_2, n_3) = (1, 1, 3)$ .

Calculating the constant term:

\text{Constant term} = \frac{5!}{1! 1! 3!} \cdot 2^1 \cdot 3^3

= \frac{120}{6} \times 2 \times 27

= 20 \times 54

= 1080