If $1^2 \cdot\left({ }^{15} C_1\right)+2^2 \cdot\left({ }^{15} C_2\right)+3^2 \cdot\left({ }^{15} C_3\right)+\ldots+15^2 \cdot\left({ }^{15} C_{15}\right)=2^m \cdot 3^n \cdot 5^k$, where $m, n, k \in \mathbf{N}$, then $\mathrm{m}+\mathrm{n}+\mathrm{k}$ is equal to :
Solution
<p>$\sum_{r=1}^{15} r^2 \cdot{ }^{15} C_r \quad\left(r^n C_r=n^{n-1} C_{r-1}\right)$</p>
<p>$$\begin{aligned}
& =15 \sum_{r=1}^{15} r \cdot{ }^{14} C_{r-1} \\
& =15 \sum_{r=1}^{15}(r-1+1){ }^{14} C_{r-1} \\
& =15 \cdot \sum_{r=1}^{15}(r-1){ }^{14} C_{r-1}+15 \cdot \sum_{r=1}^{15}{ }^{14} C_{r-1} \\
& =15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14} \\
& =15 \cdot 2^{14}(7+1) \\
& =5 \cdot 3 \cdot 2^{17} \\
& n+m+k=17+1+1=19
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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