Q49 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
Let $[\cdot]$ denote the greatest integer function and $f(x)=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}^{3}} \sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{x}}\right]$. Then $12 \sum_{\mathrm{j}=1}^{\infty} f(\mathrm{j})$ is equal to $\_\_\_\_$.
Correct answer: 2
Solution
Let [·] denote the greatest integer function. We have $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n [\frac{k^2}{3^x}]$.
For large $n$: $\sum_{k=1}^n [\frac{k^2}{3^x}] \approx \sum_{k=1}^n \frac{k^2}{3^x} = \frac{1}{3^x} \cdot \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3 \cdot 3^x}$
Therefore: $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \cdot \frac{n^3}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$
Computing the infinite series:
$\sum_{j=1}^\infty f(j) = \sum_{j=1}^\infty \frac{1}{3^{j+1}} = \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots$
$= \frac{1}{9} \cdot (1 + \frac{1}{3} + \frac{1}{9} + \cdots) = \frac{1}{9} \cdot \frac{1}{1 - 1/3} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}$
Therefore: $12 \sum_{j=1}^\infty f(j) = 12 \cdot \frac{1}{6} = 2$
For large $n$: $\sum_{k=1}^n [\frac{k^2}{3^x}] \approx \sum_{k=1}^n \frac{k^2}{3^x} = \frac{1}{3^x} \cdot \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3 \cdot 3^x}$
Therefore: $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \cdot \frac{n^3}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$
Computing the infinite series:
$\sum_{j=1}^\infty f(j) = \sum_{j=1}^\infty \frac{1}{3^{j+1}} = \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots$
$= \frac{1}{9} \cdot (1 + \frac{1}{3} + \frac{1}{9} + \cdots) = \frac{1}{9} \cdot \frac{1}{1 - 1/3} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}$
Therefore: $12 \sum_{j=1}^\infty f(j) = 12 \cdot \frac{1}{6} = 2$
Hence, the correct answer is 2.