Mathematics - Binomial Theorem Question with Solution | TestHub

In the binomial expansion of
${\left(1+x\right)}^{99}$, the sum of odd coefficients of $x$ is given by

$K=C_1+C_3+\dots .+C_{99}=\frac{2^{99}-0}{2}=2^{98}$

Also, $a=$Middle term in the binomial expansion of ${\left(2+\frac{1}{\sqrt{2}}\right)}^{200}$ is

$T_{\frac{200}{2}+1}=T_{100+1}={}^{200}C_{100}{\left(2\right)}^{100}{\left(\frac{1}{\sqrt{2}}\right)}^{100}$

$\Rightarrow T_{100+1}={}^{200}C_{100}·2^{50}$

So,

$\frac{{}^{200}C_{99}K}{a}$

$=\frac{{}^{200}C_{99}\times 2^{98}}{{}^{200}C_{100}\times 2^{50}}=\frac{100}{101}\times 2^{48}$

$=\frac{25}{101}\times 2^{50}=\left(\frac{m}{n}\right)2^l$

On comparing, we get

$l=50, m=25 \mathrm{and} n=101$

Therefore, $\left(l, n\right)\equiv \left(50,101\right)$.