JEE Mains · Maths · STD 11 - 7. binomial theoram
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 :
- A \(19\)
- B \(21\)
- C \(18\)
- D \(20\)
Answer & Solution
Correct Answer
(A) \(19\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \sum_{\mathrm{r}=1}^{15} \mathrm{r}^2\left({ }^{15} \mathrm{C}_{\mathrm{r}}\right) \Rightarrow 15 \sum_{\mathrm{r}=1}^{15} \mathrm{r}^{14} \mathrm{C}_{\mathrm{r}-1} \\ & 15 \sum_{\mathrm{r}=1}^{15}(\mathrm{r}-1+1){ }^{14} \mathrm{C}_{\mathrm{r}-1} \\ & 15 \cdot…
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