JEE Main 2025 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 4 April 2025, Shift 2, Maths Q7

Question 7 of 75 in this shift, Maths question 7 of 25, Section A.

1^2\cdot({}^{15}C_1)+2^2\cdot({}^{15}C_2)+3^2\cdot({}^{15}C_3)+\ldots+15^2\cdot({}^{15}C_{15})=2^m\cdot3^n\cdot5^k

, where

m,n,k\in\mathbf{N}

, then

m+n+k

is equal to :

(1)18
(2)19Official answer
(3)20
(4)21

Official answer

Option 2

NTA final key.

Chapter: Binomial Theorem
Topic: Binomial Coefficients
Idea tested: Second Derivative Method for Sums

Same idea in other shifts

Asked 3× in all

8 Apr 2024, Shift 1 · Q24Let $\alpha=\sum_{r=0}^{n}(4r^2+2r+1)\,{}^nC_r$ and $\beta=\left(\sum_{r=0}^{n}\frac{{}^nC_r}{r+1}\right)+\frac{1}{n+1}$ . If…HardNumerical value
26 Aug 2021, Shift 1 · Q66If $^{20}C_r$ is the co-efficient of $x^r$ in the expansion of $(1+x)^{20}$ , then the value of $\sum_{r=0}^{20} r^2\,{}^{20}C_r$ is equal…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.