Question

If$\begin{array}{l}1 + \left(2 + ^{49}C_1 + ^{49}C_2 + … ^{49}C_{49}\right) \left(^{50}C_2 + ^{50}C_4 + … ^{50}C_{50}\right) \end{array}$is equal to 2n. m , where m is odd, then n + m is equal to ______.

Answer 1

The Correct answer is 99
$1 + (1 + 2^{49})(2^{49} -1) = 2^{98}$
m = 1 , n = 98
m + n = 99