Question:

Let \(m,n\in\mathbb{N}\) and \(\gcd(2,n)=1\). If \(30\binom{30}{0}+29\binom{30}{1}+\cdots+2\binom{30}{28}+1\binom{30}{29}=n\cdot2^m\), then \(n+m\) is equal to __________.
(Here \(\binom{n}{k}={}^nC_k\))

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The sum is \(\sum_{k=0}^{30}(30-k)\binom{30}{k}\); use \(\sum k\binom{30}{k}=30\cdot2^{29}\).
Updated On: Nov 4, 2026
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Correct Answer: 45

Solution and Explanation

The sum is \(S=\sum_{k=0}^{30}(30-k)\binom{30}{k}\) (the \(k=30\) term is 0).
\(S=30\sum\binom{30}{k}-\sum k\binom{30}{k}=30\cdot2^{30}-30\cdot2^{29}=30\cdot2^{29}\).
\(30\cdot2^{29}=15\cdot2\cdot2^{29}=15\cdot2^{30}\), with \(n=15\) odd (\(\gcd(2,15)=1\)) and \(m=30\).
\(n+m=45\).
Answer: 45
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