Question

If $\left({ }^{40} C _{0}\right)+\left({ }^{41} C _{1}\right)+\left({ }^{42} C _{2}\right)+\ldots+\left({ }^{60} C _{20}\right)=\frac{ m }{ n }{ }^{60} C _{20}, m$ and $n$ are coprime, then $m + n$ is equal to______.

Answer 1

${ }^{40} C _{0}+{ }^{41} C _{1}+{ }^{42} C _{2}+\ldots \ldots{ }^{59} C _{19}+{ }^{60} C _{20}$
$\left(\frac{1}{41}+1\right){ }^{41} C _{1}+{ }^{42} C _{2}+\ldots \ldots$
${\left[\frac{42}{41}\left(\frac{2}{42}\right)+1\right]{ }^{42} C _{2}+{ }^{43} C _{3}+\ldots . . }$
$\left(\frac{2}{41}+1\right){ }^{42} C _{2}+{ }^{43} C _{3}+\ldots \ldots$
$\left(\frac{43}{41} \times \frac{3}{43}+1\right){ }^{43} C _{3}+{ }^{44} C _{4}+\ldots \ldots .$
$\frac{3+41}{41}{ }^{43} C _{3}+\ldots \ldots .$
Similarly :
$\frac{20+41}{41}$
$\Rightarrow m =61 ; n =41$
$m + n =102$