Exercise 4.4 — Question 8

Problem Statement

Find the sum of the series using sigma notation and standard summation formulas:

${\sum }_{r=1}^{n}r(r+1)(r+2)$

Solution

Expand the general term $T_r=r(r+1)(r+2)$:

$T_r=r(r^2+3r+2)=r^3+3r^2+2r$

Now sum both sides from $r=1$ to $n$:

$S_n={\sum }_{r=1}^n{T}_r={\sum }_{r=1}^n{r}^3+3{\sum }_{r=1}^n{r}^2+2{\sum }_{r=1}^{n}r$

Apply the standard summation formulas:

Substituting:

$S_n={\left[\frac{n(n+1)}{2}\right]}^2+3\cdot \frac{n(n+1)(2n+1)}{6}+2\cdot \frac{n(n+1)}{2}$

$S_n=\frac{n^2(n+1{)}^2}{4}+\frac{n(n+1)(2n+1)}{2}+n(n+1)$

Factor out $\frac{n(n+1)}{4}$:

$S_n=\frac{n(n+1)}{4}\left[n(n+1)+2(2n+1)+4\right]$

$S_n=\frac{n(n+1)}{4}\left[n^2+n+4n+2+4\right]$

$S_n=\frac{n(n+1)}{4}\left[n^2+5n+6\right]$

$S_n=\frac{n(n+1)(n+2)(n+3)}{4}$

Key Formulas Used

${\sum }_{r=1}^{n}r=\frac{n(n+1)}{2}$

${\sum }_{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6}$

${\sum }_{r=1}^n{r}^3=\frac{n^2(n+1{)}^2}{4}$