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(C) The sum of coefficients of odd powers of $x$ in $(1+x)^{99}$ is $K = 2^{99-1} = 2^{98}$.The middle term $a$ in the expansion of $(2+\frac{1}{\sqrt

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$(50, 101)$

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(C) The sum of coefficients of odd powers of $x$ in $(1+x)^{99}$ is $K = 2^{99-1} = 2^{98}$.The middle term $a$ in the expansion of $(2+\frac{1}{\sqrt{2}})^{200}$ is the $101^{st}$ term:$a = {}^{200}C_{100} (2)^{100} (\frac{1}{\sqrt{2}})^{100} = {}^{200}C_{100} \cdot 2^{100} \cdot 2^{-50} = {}^{200}C_{100} \cdot 2^{50}$.Now,consider the expression $\frac{{}^{200}C_{99} K}{a} = \frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}}$.Using the property ${}^{n}C_{r} = \frac{n-r+1}{r} {}^{n}C_{r-1}$,we have $\frac{{}^{200}C_{99}}{{}^{200}C_{100}} = \frac{100}{200-100+1} = \frac{100}{101}$.Thus,$\frac{{}^{200}C_{99} K}{a} = \frac{100}{101} \cdot 2^{98-50} = \frac{100}{101} \cdot 2^{48} = \frac{25 \cdot 4}{101} \cdot 2^{48} = \frac{25 \cdot 2^2 \cdot 2^{48}}{101} = \frac{2^{50} \cdot 25}{101}$.Comparing this with $\frac{2^{\ell} m}{n}$,we get $\ell = 50$,$m = 25$,and $n = 101$.Since $m=25$ and $n=101$ are odd,the ordered pair $(\ell, n)$ is $(50, 101)$.

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