JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
If $1^{2}\cdot\left({}^{15}C_1\right)+2^{2}\cdot\left({}^{15}C_2\right)+3^{2}\cdot\left({}^{15}C_3\right)+\ldots+15^{2}\cdot\left({}^{15}C_{15}\right)=2^{m}\cdot3^{n}\cdot5^{k}$, where $m,n,k\in\mathbb{N}$, then $m+n+k$ is equal to :
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. Use the standard sum $\sum k^2\,{}^{n}C_k$. Formulas used. (1) $\sum_{k=0}^{n}k^2\,{}^{n}C_k=n(n+1)\,2^{n-2}$Step 1 - Apply (1) with $n=15$. $$\sum_{k=1}^{15}k^2\,{}^{15}C_k=15\cdot16\cdot2^{13}.$$ Step 2 - Factorise. $$15\cdot16\cdot2^{13}=3\cdot5\cdot2^4\cdot2^{13}=2^{17}\cdot3^1\cdot5^1.$$ So $m=17$, $n=1$, $k=1$. Step 3 - Value. $m+n+k=19$. Answer: (B)