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Friday, 9 October

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.

If 12⋅(15C1)+22⋅(15C2)+32⋅(15C3)+…+152⋅(15C15)=2m⋅3n⋅5k1^2\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∈Nm,n,k\in\mathbf{N}, then m+n+km+n+k is equal to :
  1. (1)18
  2. (2)19Official answer
  3. (3)20
  4. (4)21

Official answer

Option 2

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 8 Apr 2024, Shift 1 · Q24Let α=∑r=0n(4r2+2r+1) nCr\alpha=\sum_{r=0}^{n}(4r^2+2r+1)\,{}^nC_r and β=(∑r=0nnCrr+1)+1n+1\beta=\left(\sum_{r=0}^{n}\frac{{}^nC_r}{r+1}\right)+\frac{1}{n+1}. If…HardNumerical value

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 42 of the 174 JEE Main shifts held from 2017 to 2026.