If $1+\left(2+{}^{49}\mathrm{C}_1+{}^{49}\mathrm{C}_2+\dots .+{}^{49}\mathrm{C}_{49}\right)\left({}^{50}\mathrm{C}_2+{}^{50}\mathrm{C}_4+\dots ..+{}^{50}\mathrm{C}_{50}\right)$ is equal to $2^n.m$, where $m$ is odd, then $n+m$ is equal to _____ .

Let $I=1+\left(1+{}^{49}C_0+{}^{49}C_1+\dots +{}^{49}C_{49}\right)\left({}^{50}C_2+{}^{50}C_4+\dots +{}^{50}C_{50}\right)$

As ${}^{49}C_0+{}^{49}C_1+\dots +{}^{49}C_{49}=2^{49}$

and ${}^{50}C_0+{}^{50}C_2+\dots +{}^{50}C_{50}=2^{49}$

$\Rightarrow {}^{50}C_2+{}^{50}C_4+\dots .+{}^{50}C_{50}=2^{49}-1$

$\therefore I=1+\left(2^{49}+1\right)\left(2^{49}-1\right)$$=2^{98}$

Now comparing $I=2^{n}m$ we get, $m=1$ and $n=98$

So, $m+n=99$