JEE Main 2026 — Mathematics Calculus

Let [·] denote the greatest integer function. We have $\displaystyle f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n [\frac{k^2}{3^x}]$f(x)=n→∞lim​n31​k=1∑n​[3xk2​].

For large $n$n: $\displaystyle \sum_{k=1}^n [\frac{k^2}{3^x}] \approx \sum_{k=1}^n \frac{k^2}{3^x} = \frac{1}{3^x} \cdot \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3 \cdot 3^x}$k=1∑n​[3xk2​]≈k=1∑n​3xk2​=3x1​⋅6n(n+1)(2n+1)​≈3⋅3xn3​

Therefore: $\displaystyle f(x) = \lim_{n \to \infty} \frac{1}{n^3} \cdot \frac{n^3}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$f(x)=n→∞lim​n31​⋅3⋅3xn3​=3x+11​

Computing the infinite series:

$$\sum_{j=1}^\infty f(j) = \sum_{j=1}^\infty \frac{1}{3^{j+1}} = \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots$$j=1∑∞​f(j)=j=1∑∞​3j+11​=321​+331​+341​+⋯

$$= \frac{1}{9} \cdot (1 + \frac{1}{3} + \frac{1}{9} + \cdots) = \frac{1}{9} \cdot \frac{1}{1 - 1/3} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}$$=91​⋅(1+31​+91​+⋯)=91​⋅1−1/31​=91​⋅23​=61​

Therefore: $\displaystyle 12 \sum_{j=1}^\infty f(j) = 12 \cdot \frac{1}{6} = 2$12j=1∑∞​f(j)=12⋅61​=2