Question

Find the sum of the series $1·n+2(n-1)+3(n-2)+...+(n-1)2+n·1$

Answer 1

Step 1: Given

Let $r$^th term be $T_r$

$$\begin{gathered} T_r=r[n-(r-1)] \\ =rn-r^2-r \end{gathered}$$

Step 2: Sum $T_r$ by sigma method

Let $S_n$ be the sum of the series up to $n$ terms

$$\begin{gathered} \Rightarrow S_n=\sum _{r=1}^n{T}_r \\ \Rightarrow S_n=\sum _{r=1}^n(rn-r^2-r) \\ \Rightarrow S_n=\underset{r=1}{\overset{n}{n\sum r-}}\underset{r=1}{\overset{n}{\sum r^2}}-\underset{r=1}{\overset{n}{\sum r}} \\ \Rightarrow S_n=\frac{n\left[n(n+1)\right]}{2}-\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\left[\because \sum _{k=1}^{n}k=\frac{k\left(k+1\right)}{2},\sum _{k=1}^n{k}^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left[n-\frac{\left(2n+1\right)}{3}-1\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left[\frac{3n-2n-1-3}{3}\right] \\ \Rightarrow S_n=\frac{n\left(n+1\right)\left(n-4\right)}{6} \end{gathered}$$

Therefore, $1·n+2(n-1)+3(n-2)+...+(n-1)2+n·1=\frac{n\left(n+1\right)\left(n-4\right)}{6}$