If 1^2 cdot binom{15}{1} + 2^2 cdot binom{15} | vedclass

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(A) The given sum is $S = \sum_{r=1}^{15} r^2 \binom{15}{r}$.Using the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$, we have $S = \sum_{r=1}^{15} r

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$19$

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(A) The given sum is $S = \sum_{r=1}^{15} r^2 \binom{15}{r}$.Using the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$, we have $S = \sum_{r=1}^{15} r \cdot 15 \binom{14}{r-1} = 15 \sum_{r=1}^{15} (r-1+1) \binom{14}{r-1}$.$S = 15 \left[ \sum_{r=1}^{15} (r-1) \binom{14}{r-1} + \sum_{r=1}^{15} \binom{14}{r-1} \right]$.Using $r \binom{n}{r} = n \binom{n-1}{r-1}$ again, $S = 15 \left[ 14 \sum_{r=2}^{15} \binom{13}{r-2} + 2^{14} \right]$.$S = 15 \left[ 14 \cdot 2^{13} + 2^{14} \right] = 15 \cdot 2^{13} (14 + 2) = 15 \cdot 2^{13} \cdot 16$.$S = (3 \cdot 5) \cdot 2^{13} \cdot 2^4 = 2^{17} \cdot 3^1 \cdot 5^1$.Comparing with $2^m \cdot 3^n \cdot 5^k$, we get $m=17, n=1, k=1$.Thus, $m+n+k = 17+1+1 = 19$.

If $1^2 \cdot \binom{15}{1} + 2^2 \cdot \binom{15}{2} + 3^2 \cdot \binom{15}{3} + \ldots + 15^2 \cdot \binom{15}{15} = 2^m \cdot 3^n \cdot 5^k$, where $m, n, k \in N$, then $m + n + k$ is equal to:

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