4.7 Q-2

Exercise 4.7 — Question 2

This question involves arithmetic-geometric sequences and series, combining arithmetic and geometric progressions.

---

Key Concepts

Arithmetic-Geometric Sequence

An arithmetic-geometric sequence is formed by multiplying the 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

The sum of the first $n$ terms of an arithmetic-geometric series is found using the multiply-and-subtract 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 $|r|<1$, the series converges and:

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

---

Summation Formulas (Sigma Notation)

For use with M-11-A-26:

---

Worked Example

Find the sum to infinity of the series: $1+2\cdot \frac{1}{2}+3\cdot {\left(\frac{1}{2}\right)}^2+4\cdot {\left(\frac{1}{2}\right)}^3+\cdots$

Solution:

Here $a=1$, $d=1$, $r=\frac{1}{2}$, and $|r|<1$ so the series converges.

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

$=\frac{1}{\frac{1}{2}}+\frac{\frac{1}{2}}{\frac{1}{4}}=2+2=4$