4.4 Q-9

Exercise 4.4 — Question 9

Problem Statement

Find the sum of the series whose $n$th term is $n(n+1)(n+2)$, i.e., evaluate:

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

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Solution

Expand the general term:

$T_r=r(r+1)(r+2)=r^3+3r^2+2r$

Therefore:

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

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

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

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

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

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

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Result

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

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Verification (n = 3)

$T_1=1\cdot 2\cdot 3=6,T_2=2\cdot 3\cdot 4=24,T_3=3\cdot 4\cdot 5=60$

$S_3=6+24+60=90$

Using the formula:

$S_3=\frac{3\cdot 4\cdot 5\cdot 6}{4}=\frac{360}{4}=90✓$