If 1 + 2 + C 1 49 + C 2 49 + … . + C 49 49 C 2 50 + C 4 50 + … . . + C 50 50 is equal to 2 n . m…

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, $m+n=99$