Exercise 4.8 — Question 7

Problem

Evaluate the sum using sigma notation and standard summation formulas:

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

Solution

Expand the general term:

$k(k+1)(k+2)=k^3+3k^2+2k$

Therefore:

${\sum }_{k=1}^{n}k(k+1)(k+2)={\sum }_{k=1}^n{k}^3+3{\sum }_{k=1}^n{k}^2+2{\sum }_{k=1}^{n}k$

Apply the standard summation formulas:

Substituting:

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

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

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

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

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

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

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

Final Answer

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

Key Summation Formulas Used

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

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

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