\(\mathrm{S}_1=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}, \mathrm{S}_2=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^2\) and \(\mathrm{S}_3=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^3\), then the value of \(\lim _{n \rightarrow \infty} \frac{s_1\left(1+\frac{s_3}{4}\right)}{s_2^2}\) is

\(\begin{aligned} & \mathrm{S}_1=\sum \mathrm{r}=\frac{\mathrm{r}(\mathrm{r}+1)}{2} \\ & \mathrm{~S}_2=\sum \mathrm{r}^2=\frac{\mathrm{r}(\mathrm{r}+1)(2 \mathrm{r}+1)}{6} \\ & \text { and } \mathrm{S}_3=\sum \mathrm{r}^3=\left(\frac{\mathrm{r}(\mathrm{r}+1)}{2}\right)^2\end{aligned}\)
\(\begin{aligned} \therefore \quad & \lim _{r \rightarrow \infty} \frac{S_1\left(1+\frac{S_3}{4}\right)}{S_2^2} \\ = & \lim _{\mathrm{r} \rightarrow \infty} \frac{\frac{r(r+1)}{2}\left(1+\frac{r^2(r+1)^2}{16}\right)}{\frac{r^2(r+1)^2(2 r+1)^2}{36}}\end{aligned}\)
\(=\lim _{r \rightarrow \infty} \frac{r(r+1)}{2}\left(1+\frac{r^2(r+1)^2}{16}\right) \div \frac{r^2(r+1)^2(2 r+1)^2}{36}\)
\(=\lim _{\mathrm{r} \rightarrow \infty} \frac{\mathrm{r}(\mathrm{r}+1)}{2}\left(1+\frac{\mathrm{r}^2(\mathrm{r}+1)^2}{16}\right) \times \frac{36}{\mathrm{r}^2(\mathrm{r}+1)^2(2 \mathrm{r}+1)^2}\)
\(=\lim _{\mathrm{r} \rightarrow \infty} 1\left(1+\frac{\mathrm{r}^2(\mathrm{r}+1)^2}{16}\right) \times \frac{18}{\mathrm{r}(\mathrm{r}+1)(2 \mathrm{r}+1)^2}\)
\(\begin{aligned} & =18 \lim _{\mathrm{r} \rightarrow \infty}\left(\frac{16+\mathrm{r}^2(\mathrm{r}+1)^2}{16}\right) \cdot \frac{1}{\mathrm{r}(\mathrm{r}+1)(2 \mathrm{r}+1)^2} \\ & =\frac{18}{16} \lim _{\mathrm{r} \rightarrow \infty}\left(\frac{16}{\mathrm{r}^4}+\left(1+\frac{1}{\mathrm{r}}\right)^2\right) \cdot \frac{1}{(1)\left(1+\frac{1}{r}\right)\left(2+\frac{1}{r}\right)^2}\end{aligned}\)
\(\begin{aligned} & =\frac{18}{16}\left(0+(1+0)^2\right)^2\left(\frac{1}{1(1+0)(2+0)^2}\right) \\ & =\frac{18}{16}(1) \times\left(\frac{1}{4}\right) \\ & =\frac{18}{64}=\frac{9}{32}\end{aligned}\)