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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Binomial Theorem

If (1^2 ·( ^ 15 C_1)+2^2 ·( ^ 15 C_2)+3^2 ·( ^ 15 C_3)+ ) (+15^2 ·(...

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 ⋅((1515))=\cdot\left(\binom{15}{15}\right)=2^m ⋅\cdot3^n ⋅\cdot5^k), where (m, n, k \in \mathbf{N}), then (m+n+k\mathrm{m}+\mathrm{n}+\mathrm{k}) is equal to :

Options

  1. A1919
  2. B2121
  3. C1818
  4. D2020

Correct answer

A. 1919

Step-by-step solution

∑r=115r2(15Cr)⇒15∑r=115r14Cr−115∑r=115(r−1+1)14Cr−115⋅14∑r=11513Cr−2+15∑r=11514Cr−115⋅14⋅213+15⋅21431⋅213(70+10)31⋅213⋅8031⋅51⋅217m=17n=1k=1 option (1)\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 14 \sum_{\mathrm{r}=1}^{15}{ }^{13} \mathrm{C}_{\mathrm{r}-2}+15 \sum_{\mathrm{r}=1}^{15}{ }^{14} \mathrm{C}_{\mathrm{r}-1} \\ & 15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14} \\ & 3^1 \cdot 2^{13}(70+10) \\ & 3^1 \cdot 2^{13} \cdot 80 \\ & 3^1 \cdot 5^1 \cdot 2^{17} \\ & \mathrm{ m}=17 \mathrm{n}=1 \quad \mathrm{k}=1 \\ & \text{ option }(1)\end{aligned}
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