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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 $\frac{{ }^{200} C _{99} K }{ a }=\frac{2^{\ell} m }{ n }$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n )$ is equal to :
$(50,51)$
$(51,99)$
$(50,101)$
$(51,101)$
Solution
In the expansion of
$(1+ x )^{99}= C _0+ C _1 x + C _2 x ^2+\ldots+ C _{99} x ^{99}$
$K = C _1+ C _3+\ldots . .+ C _{99}=2^{98}$
$a$ Middle in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$
$\frac{T_{200}}{2}+1 ={ }^{200} C _{100}(2)^{100}\left(\frac{1}{\sqrt{2}}\right)^{100}$
$={ }^{200} C _{100} \cdot 2^{50}$
So, $\frac{{ }^{200} C _{99} \times 2^{98}}{{ }^{900} C _{100} \times 2^{50}}=\frac{100}{101} \times 2^{48}$
So, $\frac{25}{101} \times 2^{50}=\frac{ m }{ n } 2^{\prime}$
$\therefore m , n$ are odd so
$(\ell, n )$ become $(50,101)$ Ans.
