Mathematics · 2021
JEE Main · 26 February 2021, Shift 1 · Q90
Let m, n ∈ N and gcd (2, n )=1. If 30 30 0 +29 30 1 +…..+2 30 28 +1 30 29 = n · 2^m, then n + m is equal to \ \ \ \. ( Here n k =^n C_k )
Let $\displaystyle \mathrm{m}, \mathrm{n} \in \mathbf{N}$ and $\displaystyle \operatorname{gcd}(2, \mathrm{n})=1$. If $\displaystyle 30\binom{30}{0}+29\binom{30}{1}+\ldots . .+2\binom{30}{28}+1\binom{30}{29}=\mathrm{n} \cdot 2^{\mathrm{m}}$, then $\displaystyle \mathrm{n}+\mathrm{m}$ is equal to \_\_\_\_.
$$\left(\text { Here }\binom{\mathrm{n}}{\mathrm{k}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{k}}\right)
$$
Official answer
45
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.