Question

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

Answer 1

We have,

${lim}_{x\to \infty }\frac{1.n+2\left(n-1\right)+3\left(n-2\right)+....+n.1}{n^3}{lim}_{x\to \infty }\frac{{\sum }_{r=1}^{n}r\left(n-r+1\right)}{n^3}$

${lim}_{x\to \infty }\frac{n\times \frac{n\left(n+1\right)}{2}+\frac{n\left(n+1\right)}{2}-\frac{n\left(n+1\right)\left(2n+1\right)}{6}}{n^3}{lim}_{x\to \infty }\frac{\frac{n\left(n+1\right)}{2}\left(n+1\right)-\frac{n\left(n+1\right)\left(2n+1\right)}{6}}{n^3}={lim}_{x\to \infty }\frac{\frac{n^3}{2}-\frac{n^3}{3}+......}{n^3}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

Hence, this is the answer.