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 ${(2 + \frac{1}{\sqrt{2}})}^{200}$ If $\frac{200_{c_{99}} K}{a} = \frac{2^{l} . m}{n}$ , where m and n are odd numbers then $l + n$ is equal to

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answer is 151.

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Detailed Solution

${(1 + x)}^{99}$

$K = \frac{f (1) - f (- 1)}{2} = \frac{2^{99}}{2} = 2^{98}$ \

${(2 + \frac{1}{\sqrt{2}})}^{200}$

a=middle term= $T_{\frac{200}{2} + 1}$

$a = T_{100 + 1} = 200_{C_{100}} 2^{100} . \frac{1}{2^{50}}$

$\frac{200_{C_{99}} 2^{98}}{200_{C_{100}} {.2}^{50}} = \frac{2^{l} . m}{n}$

$\frac{25. 2^{50}}{101} = \frac{2^{l} . m}{n}$

l=50, n=101
l+n=151.

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