Let [cdot]$[\cdot]$ denote the greatest integer function and f(x)=lim_ntoinftyfrac1n^3sum_k=1^nleft[frack^23^xright]$f(x)=\lim_{n\to\infty}\frac{1}{n^{3}}\sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right]$. Then 12sum_j=1^inftyf(j)$12\sum_{j=1}^{\infty}f(j)$ is equal to

### Related Formula textSandwich Theorem for greatest integer function: x - 1 < [x] leq x$$\text{Sandwich Theorem for greatest integer function: } x - 1 < [x] \leq x$$ sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ ### Core Logic Evaluate f(x)$f(x)$ using Squeeze Theorem on the summation bounds due to the Greatest Integer Function. frack^23^x - 1 < left[frack^23^xright] leq frack^23^x$$\frac{k^2}{3^x} - 1 < \left[\frac{k^2}{3^x}\right] \leq \frac{k^2}{3^x}$$ ### Step 1: Evaluate Limit f(x) Sum over bounds: sum_k=1^n left( frack^23^x - 1 right) < sum_k=1^n left[ frack^23^x right] leq sum_k=1^n frack^23^x$$\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \sum_{k=1}^n \frac{k^2}{3^x}$$ frac13^x fracn(n+1)(2n+1)6 - n < sum_k=1^n left[ frack^23^x right] leq frac13^x fracn(n+1)(2n+1)6$$\frac{1}{3^x} \frac{n(n+1)(2n+1)}{6} - n < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \frac{1}{3^x} \frac{n(n+1)(2n+1)}{6}$$ Divide by n^3$n^3$ and apply limit n to infty$n \to \infty$: lim_n to infty left( frac2n^3 + 3n^2 + n6n^3 cdot 3^x - frac1n^2 right) < f(x) leq lim_n to infty frac2n^3 + 3n^2 + n6n^3 cdot 3^x$$\lim_{n \to \infty} \left( \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x} - \frac{1}{n^2} \right) < f(x) \leq \lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x}$$ f(x) = frac26 cdot 3^x = frac13 cdot 3^x = frac13^x+1$$f(x) = \frac{2}{6 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$$ ### Step 2: Evaluate Final Summation We need 12sum_j=1^infty f(j)$12\sum_{j=1}^{\infty} f(j)$: 12 sum_j=1^infty frac13^j+1 = 12 left( frac13^2 + frac13^3 + ldots right)$$12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{3^2} + \frac{1}{3^3} + \ldots \right)$$ This is an infinite geometric progression with a = frac19$a = \frac{1}{9}$ and r = frac13$r = \frac{1}{3}$. textSum = fraca1 - r = frac1/91 - 1/3 = frac1/92/3 = frac16$$\text{Sum} = \frac{a}{1 - r} = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{6}$$ Finally, 12 times frac16 = 2$12 \times \frac{1}{6} = 2$. ### Pattern Recognition When evaluating infinite limits over greatest integer sums lim_n to infty frac1n^p+1 sum [k^p/C]$\lim_{n \to \infty} \frac{1}{n^{p+1}} \sum [k^p/C]$, the G.I.F brackets can be safely replaced by exact values due to Sandwich theorem bounding the -1$-1$ residual to 0$0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits, Continuity and Differentiability Class 11 Maths: Sequence and Series