If 1 2 × ( 15 C 1 )+2 2 × ( 15 C 2 )+3 2 × ( 15 C 3 )+ldots+15 2 × ( 15 C 15 ) =2 m .3 n . 5 k , where m, n, k in mathbb N , then m+n+k is equal to :

Mathematics · Binomial Theorem

JEE Main 2025 — 4 April, Evening Shift — Question 23

If $1^{2} \cdot\left({ }^{15} C_{1}\right)+2^{2} \cdot\left({ }^{15} C_{2}\right)+3^{2} \cdot\left({ }^{15} C_{3}\right)+\ldots+15^{2} \cdot\left({ }^{15} C_{15}\right)$ $=2^{m} .3^{n}$ . $5^{k}$ , where

$m, n, k \in \mathbb{N}$ , then $m+n+k$ is equal to :

Option A:
18
Option B:
19
Correct
Option C:
21
Option D:
20

Answer: B

Step-by-step solution

$\sum_{r=1}^{15} r^{2} \cdot{ }^{15} C_{r} \quad\left(r^{n} C_{r}=n^{n-1} C_{r-1}\right)$

$=15 \sum_{r=1}^{15} r \cdot{ }^{14} C_{r-1}$

$=15 \sum_{r=1}^{15}(r-1+1){ }^{14} C_{r-1}$

$=15 \cdot \sum_{r=1}^{15}(r-1){ }^{14} C_{r-1}+15 \cdot \sum_{r=1}^{15}{ }^{14} C_{r-1}$

$=15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14}$

$=15.2^{14}(7+1)$

$=5 \cdot 3 \cdot 2^{17}$

$n+m+k=17+1+1=19$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Series involving sum & product of Binomial Coefficients