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
$(1+x{)}^{99}=C_0+C_{1}x+C_2{x}^2+\dots .+C_{99}{x}^{99}$
$K=C_1+C_3+\dots .+C_{99}=2^{98}$
$a\Rightarrow$ 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 $(50,101)$