Let m, n in mathbb{N} and operatorname{gcd}(2 | vedclass

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Question

(A) The given expression is $S = \sum_{k=0}^{29} (30-k) \binom{30}{k}$.Using the property $\binom{n}{k} = \binom{n}{n-k}$,we can rewrite the sum as:$S

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) The given expression is $S = \sum_{k=0}^{29} (30-k) \binom{30}{k}$.Using the property $\binom{n}{k} = \binom{n}{n-k}$,we can rewrite the sum as:$S = \sum_{k=0}^{29} (30-k) \binom{30}{30-k}$.Let $r = 30-k$. As $k$ goes from $0$ to $29$,$r$ goes from $30$ to $1$.$S = \sum_{r=1}^{30} r \binom{30}{r}$.Using the identity $\sum_{r=1}^{n} r \binom{n}{r} = n 2^{n-1}$:$S = 30 \cdot 2^{30-1} = 30 \cdot 2^{29}$.$S = 15 \cdot 2 \cdot 2^{29} = 15 \cdot 2^{30}$.Given $S = n \cdot 2^m$ with $\operatorname{gcd}(2, n) = 1$,we have $n = 15$ and $m = 30$.Therefore,$n + m = 15 + 30 = 45$.

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

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