JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 13)
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 :
20
19
18
21
Explanation
$\sum_{r=1}^{15} r^2 \cdot{ }^{15} C_r \quad\left(r^n C_r=n^{n-1} C_{r-1}\right)$
$$\begin{aligned} & =15 \sum_{r=1}^{15} r \cdot{ }^{14} C_{r-1} \\ & =15 \sum_{r=1}^{15}(r-1+1){ }^{14} C_{r-1} \\ & =15 \cdot \sum_{r=1}^{15}(r-1){ }^{14} C_{r-1}+15 \cdot \sum_{r=1}^{15}{ }^{14} C_{r-1} \\ & =15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14} \\ & =15 \cdot 2^{14}(7+1) \\ & =5 \cdot 3 \cdot 2^{17} \\ & n+m+k=17+1+1=19 \end{aligned}$$
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