Question

If $\left({}^{40}{C}_0\right)+\left({}^{41}{C}_1\right)+\left({}^{42}{C}_2\right)+\dots +\left({}^{60}{C}_{20}\right)=\frac{m}{n}{}^{60}{C}_{20},m$ and $n$ are coprime, then $m+n$ is equal to______.

Answer 1

Answer: 102

${}^{40}{C}_0+{}^{41}{C}_1+{}^{42}{C}_2+\dots \dots {}^{59}{C}_{19}+{}^{60}{C}_{20}$
$\left(\frac{1}{41}+1\right){}^{41}{C}_1+{}^{42}{C}_2+\dots \dots$
$\left[\frac{42}{41}\left(\frac{2}{42}\right)+1\right]{}^{42}{C}_2+{}^{43}{C}_3+\dots ..$
$\left(\frac{2}{41}+1\right){}^{42}{C}_2+{}^{43}{C}_3+\dots \dots$
$\left(\frac{43}{41}\times \frac{3}{43}+1\right){}^{43}{C}_3+{}^{44}{C}_4+\dots \dots .$
$\frac{3+41}{41}{}^{43}{C}_3+\dots \dots .$
Similarly :
$\frac{20+41}{41}$
$\Rightarrow m=61;n=41$
$m+n=102$