If 1+(2+^49C_1+^49C_2++^49C_49)(^50C_2+^50C_4++^50C_50) is equal to 2^n m, where m is odd,…

Question

If $1+\left(2+{}^{49}C_1+{}^{49}C_2+\cdots+{}^{49}C_{49}\right)\left({}^{50}C_2+{}^{50}C_4+\cdots+{}^{50}C_{50}\right)$ is equal to $2^n\cdot m$, where $m$ is odd, then $n+m$ is equal to \_\_\_\_\_\_\_.

Answer: 99

Accepted Answer

$${}^{49}C_0+{}^{49}C_1+\cdots+{}^{49}C_{49}=2^{49}\quad\Rightarrow\quad{}^{50}C_2+\cdots+{}^{50}C_{50}=2^{49}-1$$

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

So $m=1,\ n=98$, giving $n+m=99$.

Solutions · 1