4.6 Q-2

Exercise 4.6 — Question 2

This question typically involves evaluating series using standard summation formulas or working with arithmetic-geometric sequences.

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 the general term is:

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

Sum to $n$ terms (using the subtraction method):

$S_n=\frac{a-[a+(n-1)d]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2},r\ne 1$

Sum to infinity (when $|r|<1$):

$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

Sigma Notation

Series can be written compactly using sigma notation. For example:

$1+2r+3r^2+4r^3+\cdots +n{r}^{n-1}={\sum }_{k=1}^{n}k{r}^{k-1}$

Worked Approach

To evaluate a series in Exercise 4.6 Q2:

1. Identify whether the series is a pure summation (∑n, ∑n², ∑n³) or an arithmetic-geometric series.
2. Apply the appropriate formula.
3. Simplify the result.