$\mathrm{S}_1=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}, \mathrm{S}_2=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^2$ 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

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Aicharya · Accepted Answer

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

$\mathrm{S}_1=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}, \mathrm{S}_2=\sum_{\mathrm{r}=1}^{\mathrm{n}}…

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