Exercise 4.8 — Question 6

Topic: Summation Formulas and Arithmetic-Geometric Series

This question applies the standard summation formulas and/or arithmetic-geometric series techniques from Exercise 4.8.

Key Summation Formulas

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

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

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

Arithmetic-Geometric Series

An arithmetic-geometric sequence is formed by multiplying corresponding terms of an arithmetic sequence and a geometric sequence.

If the arithmetic sequence has first term $a$ and common difference $d$, and the geometric sequence has first term $1$ and common ratio $r$, then:

General term: $T_n=[a+(n-1)d]r^{n-1}$

Sum to $n$ terms (using the subtraction method, $r\ne 1$): $S_n=\frac{a-[a+(n-1)d]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2}$

Sum to infinity (when $|r|<1$): $S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

Worked Approach for Q6

1. Identify whether the series is a pure summation (∑n, ∑n², ∑n³) or an arithmetic-geometric series.
2. Write the general term $T_k$ in sigma notation.
3. Apply the appropriate formula.
4. Simplify the result.