If 1+left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+ldots .+{ }^{49} C _{49}right)left({ }^{50} C _{2}+{ }^{

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7.Binomial Theorem

hard

If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

A

$98$

B

$97$

C

$96$

D

$99$

(JEE MAIN-2022)

Solution

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

$m=1, n=98$

$m + n =99$

Standard 11

Mathematics

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