Let [ ] denote the greatest integer function and f(x)=n lim1n^3 _k=1^n[k^23^x].

Question

Let $[\cdot ]$ denote the greatest integer function and $f(x)=\underset{n	o \infty}{lim}\frac{1}{{n}^{3}}\sum _{k=1}^{n}[\frac{{k}^{2}}{{3}^{x}}]$. Then $12\sum _{j=1}^{\infty}f(i)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

$\begin{gathered} f(x)=\underset{x	o \infty}{lim}(\frac{1}{{n}^{3}}\cdot \sum _{k=1}^{n}[\frac{{k}^{2}}{{3}^{x}}])=\underset{n	o \infty}{lim}\frac{1}{{n}^{3}}\cdot \sum _{k=1}^{n}(\frac{{k}^{2}}{{3}^{x}})-\underset{n	o \infty}{lim}\frac{1}{{n}^{3}}\cdot \sum _{k=1}^{n}{\frac{{k}^{2}}{{3}^{x}}} \ f(x)=\underset{n	o \infty}{lim}\frac{1}{{n}^{3}}	imes \frac{n(n+1)(2n+1)}{6	imes {3}^{x}}-0\Rightarrow f(x)=\frac{1}{{3}^{x+1}} \ 12\sum _{j=1}^{\infty}f(j)=12(\frac{1}{{3}^{2}}+\frac{1}{{3}^{3}}+\frac{1}{{3}^{4}}+\ldots .)=12(\frac{\frac{1}{9}}{1-\frac{1}{3}})=\frac{12}{9-3}=2 \end{gathered}$

Solutions · 1