Let $[\cdot]$ denote the greatest integer function and $f(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}^{3}} \sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right]$. Then $12 \sum_{\mathrm{j}=1}^{\infty} f(\mathrm{j})$ is equal to ____.

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Definite Integrals

Let

[\cdot]

denote the greatest integer function and

f (x) = lim_{n \to \infty} \frac{1}{n ^{3}} \sum_{k = 1}^{n} [\frac{k ^{2}}{3 ^{x}}]

. Then

12 \sum_{j = 1}^{\infty} f (j)

is equal to ____.

Answer: 2

Using the squeeze theorem on the greatest integer function:

k = 1 \sum n (\frac{k ^{2}}{3 ^{x}} - 1) < k = 1 \sum n [\frac{k ^{2}}{3 ^{x}}] \leq k = 1 \sum n \frac{k ^{2}}{3 ^{x}}

Dividing by

n^{3}

and taking limit as

n \to \infty

n \to \infty lim \frac{n ( n + 1 ) ( 2 n + 1 )}{6 n ^{3} \cdot 3 ^{x}} = \frac{1}{3 ^{x + 1}}

f (x) = \frac{1}{3 ^{x + 1}}

Therefore:

12 j = 1 \sum \infty f (j) = 12 j = 1 \sum \infty \frac{1}{3 ^{j + 1}} = 12 \cdot \frac{1/9}{1 - 1/3} = 12 \cdot \frac{1}{6} = 2

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer