JEE Main 2026 — 21 January, Evening Shift — Question 21

$\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}-1\right)<\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right] \leq \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}$

$\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6.3^{\mathrm{x}}}<\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right] \leq \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6.3^{\mathrm{x}}}$

$$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^{3} \cdot 3^{x}}<\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right] \leq \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 \cdot 3^{x} \cdot n^{3}} & \end{aligned}$$

$\quad \frac{1}{3^{x+1}}<\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right] \leq \frac{1}{3^{x+1}}$

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

$\Rightarrow 12 \sum_{j=1}^{\infty} f(j)=12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}}=12\left[\frac{1}{9}+\frac{1}{27}+--\infty\right]$

$12 \sum_{\mathrm{j}=1}^{\infty} f(\mathrm{j})$ $\quad=12\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=2$