Question

${lim}_{n\to \infty }\frac{1.n^2+2(n-1{)}^2+3(n-2{)}^2+...+n{.1}^2}{1^3+2^3+3^3+...+n^3}$

Answer 1

$\begin{array}{c}Accordintoquestion,\\ weneedsformula:\sum n=\frac{n(n+1)}{2}\\ \sum n^2=\frac{n(n+1)(2n+1)}{6}\\ \sum n^3=\frac{n{(n+1)}^2}{3}\\ Now,fromnumerator:Ingeneralterm..........\\ \Rightarrow N:1.n^2+2(n-1{)}^2+3(n-2{)}^2+...+n{.1}^2\\ {\sum }_{r=1}^n\left(r\right)(n-(r-1){)}^2\\ \Rightarrow {\sum }_{r=1}^{n}r\left[n^2+r^2-2r+1-2n(r-1)\right]\\ \Rightarrow {\sum }_{r=1}^{n}r\left[n^2+r^2-2r+1-2nr+2n\right]\\ \Rightarrow {\sum }_{r=1}^{n}r\left[r^2-2r(1+n)+n^2+1+2n\right]\\ \Rightarrow {\sum }_{r=1}^n\left[r^3-2r^2(1+n)+{(n+1)}^{2}r\right]\\ \Rightarrow {\sum }_{r=1}^n{r}^3-2(1+n){\sum }_{r=1}^n{r}^2+{(n+1)}^2{\sum }_{r=1}^{n}r\\ Now,\\ \frac{{\left(n\right)}^2{(n+1)}^2}{4}-\frac{2{(n+1)}^{2}n(2n+1)}{6}+\frac{{(n+1)}^{2+1}n}{2}\\ =(n+1{)}^2\left[\frac{n^2}{4}-\frac{n(2n-1)}{3}+\frac{n(n+1)}{2}\right]\\ =(n+1{)}^2{n}^2\left[\frac{1}{4}-\frac{1}{3}(2+\frac{1}{n})+\frac{1}{2}(1+\frac{1}{n})\right]\\ =(n+1{)}^2{n}^2\left[\frac{1}{4}-\frac{2}{3}+\frac{1}{3n})+\frac{1}{2}+\frac{1}{2n}\right]\\ And,fromDenomenator:Ingeneralterm..........\\ D:1^3+2^3+3^3+...+n^3|weknowsumofcubeis{}^{\prime }{n}^{\prime }naturalno.\\ \sum n^3=\frac{n^2{(n+1)}^2}{4}\\ Now,from:........\\ \frac{N}{D}=\\ {lim}_{n\to \infty }\Rightarrow \frac{n^2{(n+1)}^2\left[\frac{1}{4}-\frac{2}{3}+\frac{1}{3n}+\frac{1}{2}+\frac{1}{2n}\right]}{\frac{n^2{(n+1)}^2}{4}}\\ {lim}_{n\to \infty }\Rightarrow \frac{\left[\frac{1}{4}-\frac{2}{3}+\frac{1}{3n}+\frac{1}{2}+\frac{1}{2n}\right]}{\frac{1}{4}}|here:nisinity\\ \Rightarrow \frac{\frac{3}{4}-\frac{2}{3}}{\frac{1}{4}}=\frac{1}{12}\times \frac{4}{1}=\frac{1}{3}Ans.\end{array}$