If $\left({}^{40}C_0\right)+\left({}^{41}C_1\right)+\left({}^{42}C_2\right)+\dots +\left({}^{60}C_{20}\right)=\frac{m}{n}{}^{60}C_{20}$ and n are coprime, then m+n is equal to _________.

$$\begin{gathered} {}^{40}C_0+{}^{41}C_1+{}^{42}C_2+\dots ..{}^{59}C_{19}+{}^{60}C_{20} \\ \left(\frac{1}{41}+1\right){}^{41}C_1+{}^{42}C_2+\dots \dots \\ \left[\frac{42}{41}\left(\frac{2}{42}\right)+1\right]{}^{42}C_2+{}^{43}C_3+\dots .. \\ \left(\frac{2}{41}+1\right){}^{42}C_2+{}^{43}C_3+\dots .. \\ \left(\frac{43}{41}\times \frac{3}{43}+1\right){}^{43}C_3+{}^{44}C_4+\dots \dots . \\ \frac{3+41}{41}·{}^{43}C_3+\dots \dots . \end{gathered}$$

Similarly:

$$\begin{gathered} \frac{20+41}{41} \\ \Rightarrow m=61;n=41 \\ m + n =102 \end{gathered}$$