If ${}^{40}C_0+{}^{41}C_1+{}^{42}C_2+\cdots +{}^{60}C_{20}=\frac{m}{n}\times {}^{60}C_{20}$ where $m$ & $n$ are co-prime, then $m+n$ is equal to

We know ${}^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r$

Also ${}^{40}C_0={}^{41}C_0$

So ${}^{41}C_0+{}^{41}C_1+{}^{42}C_2+\dots .{}^{60}C_{20}$

$={}^{42}C_1+{}^{42}C_2+{}^{43}C_3....{}^{60}C_{20}$ $={}^{43}C_2+{}^{43}C_3+\dots {}^{60}C_{20}$

$={}^{60}C_{19}+{}^{60}C_{20}={}^{61}C_{20}$

Given ${}^{61}C_{20}=\frac{m}{n}{}^{60}C_{20}$

$\Rightarrow {}^{61}C_{41}=\frac{61}{41}{}^{60}C_{40}$

$\Rightarrow \frac{61}{41}{}^{60}C_{20}=\frac{m}{n}{}^{60}C_{20}$ 

$\Rightarrow \frac{61}{41}=\frac{m}{n} \Rightarrow m+n=102$