JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $k$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1 + x)^{99}$ . Let $a$ be the middle term in the expansion of $\left(2 + \frac{1}{\sqrt{2}}\right)^{200}$ . If $\frac{{}^{200}C_{99}k}{a} = \frac{2^l m}{n}$ , where $m$ and $n$ are odd numbers, then the ordered pair $(l, n)$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: $(1+x)^{99}$ mein odd powers ke coefficients ka sum $k = 2^{99-1} = 2^{98}$ hota hai.

Step 2: $\left(2 + \frac{1}{\sqrt{2}}\right)^{200}$ ka middle term $a = {}^{200}C_{100} (2)^{100} \left(\frac{1}{\sqrt{2}}\right)^{100} = {}^{200}C_{100} \cdot 2^{100} \cdot 2^{-50} = {}^{200}C_{100} \cdot 2^{50}$ .

Step 3: Expression: $\frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}} = \frac{100}{101} \cdot 2^{48} = \frac{2^2 \cdot 25}{101} \cdot 2^{48} = \frac{2^{50} \cdot 25}{101}$ .

Step 4: Yahan $m = 25$ aur $n = 101$ (dono odd hain), toh $l = 50$ .

Answer: $(l, n) = (50, 101)$