Exercise 4.7 — Question 6

Key Summation Formulas

Before solving, recall the three standard summation formulas:

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

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

${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$

Question

Find the sum of the series: $\sum _{k=1}^n(2k^3-k^2+3k-1)$

Solution

Split the summation using linearity:

$S=2{\sum }_{k=1}^n{k}^3-{\sum }_{k=1}^n{k}^2+3{\sum }_{k=1}^{n}k-{\sum }_{k=1}^{n}1$

Substitute the standard formulas:

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

Simplify each term:

$S=\frac{n^2(n+1{)}^2}{2}-\frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}-n$

Factor out $n$ from all terms:

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

Take LCM = 6 inside the bracket:

$S=\frac{n}{6}\left[3n(n+1{)}^2-(n+1)(2n+1)+9(n+1)-6\right]$

Expand and collect:

$=\frac{n}{6}\left[3n(n^2+2n+1)-(2n^2+3n+1)+9n+9-6\right]$

$=\frac{n}{6}\left[3n^3+6n^2+3n-2n^2-3n-1+9n+9-6\right]$

$=\frac{n}{6}\left[3n^3+4n^2+9n+2\right]$

$\boxed{S=\frac{n(3n^3+4n^2+9n+2)}{6}}$

Verification (n = 1)

Direct: $2(1{)}^3-(1{)}^2+3(1)-1=2-1+3-1=3$

Formula: $\frac{1(3+4+9+2)}{6}=\frac{18}{6}=3$ ✓