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Mathematics · 2023

JEE Main · 29 January 2023, Shift 2 · Q64

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 (2+1/(√…

Let K be the sum of the coefficients of the odd powers of $\displaystyle x$ in the expansion of $\displaystyle (1+x)^{99}$. Let $\displaystyle a$ be the middle term in the expansion of $\displaystyle \left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\displaystyle \frac{{ }^{200} \mathrm{C}_{99} \mathrm{~K}}{a}=\frac{2^l \mathrm{~m}}{\mathrm{n}}$, where m and n are odd numbers, then the ordered pair $\displaystyle (l, \mathrm{n})$ is equal to
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.