4.8 Q-12

Exercise 4.8 — Question 12

Problem

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

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

Solution

Expand the general term:

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

Apply sigma notation linearity:

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

Use the standard summation formulas:

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

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

Substitute:

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

Factor out $\frac{n(n+1)}{6}$:

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

$=\frac{n(n+1)(2n+4)}{6}$

$=\frac{n(n+1)\cdot 2(n+2)}{6}$

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

Key Formulas Used

Verification (n = 3)

Direct: $1(2)+2(3)+3(4)=2+6+12=20$

Formula: $\frac{3\cdot 4\cdot 5}{3}=\frac{60}{3}=20$ ✓