Question

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 $(\ell, n )$ is equal to :

• A. $(50,51)$
• B. $(51,99)$
• C. $(50,101)$
• D. $(51,101)$

Answer 1

Answer: (C)

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}$
$a \Rightarrow$ Middle in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$
$T _{\frac{200}{2}+1} ={ }^{200} C _{100}(2)^{100}\left(\frac{1}{\sqrt{2}}\right)^{100} $
$ ={ }^{200} C _{100} \cdot 2^{50}$
So, $\frac{{ }^{200} C _{99} \times 2^{98}}{{ }^{200} C _{100} \times 2^{50}}=\frac{100}{101} \times 2^{48}$
So, $\frac{25}{101} \times 2^{50}=\frac{m}{n} 2^{\ell}$
$\therefore m , n$ are odd so
$(\ell, n )$ become $(50,101)$