Mathematics · 2025
JEE Main · 4 April 2025, Shift 2 · Q7
If 1^2 ·(^15 C_1)+2^2 ·(^15 C_2)+3^2 ·(^15 C_3)+…+15^2 ·(^15 C_15)=2^m · 3^n · 5^k, where m, n, k ∈ N, then m + n + k is equal to:
If $\displaystyle 1^2 \cdot\left({ }^{15} \mathrm{C}_1\right)+2^2 \cdot\left({ }^{15} \mathrm{C}_2\right)+3^2 \cdot\left({ }^{15} \mathrm{C}_3\right)+\ldots+15^2 \cdot\left({ }^{15} \mathrm{C}_{15}\right)=2^{\mathrm{m}} \cdot 3^{\mathrm{n}} \cdot 5^{\mathrm{k}}$, where m, n, $\displaystyle \mathrm{k} \in \mathbf{N}$, then $\displaystyle \mathrm{m}+\mathrm{n}+\mathrm{k}$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 19$
More from Binomial Theorem
- If Σ_r=0^5 (^11 C_2 r+1)/(2 r+2)=m/n, gcd(m, n)=1, then m-n is equal to ____.2025
- If the sum of the coefficients of x^7 and x^14 in the expansion of (1/x^3-x^4)^n, x ≠ 0, is zero, then the value of n is ____.2026
- If (1/(^15 C_0)+1/(^15 C_1))(1/(^15 C_1)+1/(^15 C_2)) ⋯(1/(^15 C_12)+1/(^15 C_13))=(α^13)/(^14 C_0^14 C_1 …^14 C_12), then 30 α is equal to ____.2026
- The sum of the coefficients of x^499 and x^500 in (1+x)^1000+x(1+x)^999+x^2(1+x)^998+…+x^1000 is:2026
- Let α, β, γ and δ be the coefficients of x^7, x^5, x^3 and x respectively in the expansion of (x+√(x^3-1))^5+(x-√(x^3-1))^5, x>1. If u and v satisfy…2025
- If Σ_r=1^9((r+3)/2^r) ·^9 C_r=α(3/2)^9-β, α, β ∈ N, then (α+β)^2 is equal to2025
- The remainder when ((64)^(64))^(64) is divided by 7 is equal to2025
- For some n ≠ 10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of (1+x)^n+4 be in A.P. Then the largest coefficient in…2025
JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.