8.05 Geometric series (Enrichment)

If the first term of a geometric sequence is $u_1$u1​ and the common ratio is $r$r, then the sequence is given by:

$u_1,u_1r,u_1r^2,u_1r^3,u_1r^4,\ldots$u1​,u1​r,u1​r2,u1​r3,u1​r4,…

Summing $n$n terms in a geometric sequence

Suppose we wish to add the first $n$n terms of this sequence. This will form a geometric series. We could write the sum as:

$S_n=u_1+u_1r+u_1r^2+u_1r^3+...+u_1r^{n-1}$Sn​=u1​+u1​r+u1​r2+u1​r3+...+u1​rn−1

We saw a nifty trick for finding the formula for an arithmetic series by adding two sums together and pairing up terms. We will use a similar method here. If we multiply both sides of our geometric sum by the common ratio $r$r we see that:

$rS_n=u_1r+u_1r^2+u_1r^3+...+u_1r^{n-1}+u_1r^n$rSn​=u1​r+u1​r2+u1​r3+...+u1​rn−1+u1​rn

Then, by carefully subtracting $rS_n$rSn​ from $S_n$Sn​ term by term, we see that all of the middle terms disappear:

$S_n-rS_n=u_1+\left(u_1r-u_1r\right)+\left(u_1r^2-u_1r^2\right)+...+\left(u_1r^{n-1}-u_1r^{n-1}\right)-u_1r^n$Sn​−rSn​=u1​+(u1​r−u1​r)+(u1​r2−u1​r2)+...+(u1​rn−1−u1​rn−1)−u1​rn

This means that:

$S_n-rS_n=u_1-u_1r^n$Sn​−rSn​=u1​−u1​rn

and when common factors are taken out on both sides of this equation, we find:

$S_n\left(1-r\right)=u_1\left(1-r^n\right)$Sn​(1−r)=u1​(1−rn)

Finally, by dividing both sides by $\left(1-r\right)$(1−r) (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:

$S_n=\frac{u_1\left(1-r^n\right)}{1-r}$Sn​=u1​(1−rn)1−r​

An extra step, multiplying the numerator and denominator by $-1$−1, reveals a different form for $S_n$Sn​. Both formulas will work in any situation, particularly when using a calculator. This form is generally easier to manage when the common ratio is greater than $r=1$r=1:

$S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn​=u1​(rn−1)r−1​

The formulas for $S_n$Sn​ exclude the case for $r=1$r=1. For the case where $r=1$r=1, then the sequence becomes $u_1,u_1,u_1,u_1,\ldots$u1​,u1​,u1​,u1​,….

Is this a geometric sequence with $r=1$r=1 or an arithmetic sequence with $d=0$d=0? Either way, every term is clearly identical. Hence, the sum of the first $n$n terms is:

$S_n$Sn​	$=$=	$u_1+u_1+u_1+...+u_1$u1​+u1​+u1​+...+u1​	($n$n times)
$S_n$Sn​	$=$=	$nu_1$nu1​

Worked example

Example 1

If the sum for the first $n$n terms of the geometric sequence $5,10,20,\ldots$5,10,20,… is $5115$5115, find $n$n.

Think: We have an increasing geometric sequence. State $u_1$u1​, $r$r and $S_n$Sn​, then substitute into the formula $S_n=\frac{u_1\left(r^n-1\right)}{r-1}$Sn​=u1​(rn−1)r−1​ and rearrange.

Do: $u_1=5$u1​=5, $r=2$r=2 and $S_n=5115$Sn​=5115, so we have:

$S_n$Sn​	$=$=	$\frac{u_1\left(r^n-1\right)}{r-1}$u1​(rn−1)r−1​
$5115$5115	$=$=	$\frac{5\left(2^n-1\right)}{2-1}$5(2n−1)2−1​	Substitute values into formula
Hence,$5\left(2^n-1\right)$5(2n−1)	$=$=	$5115$5115	Simplify fraction and bring unknown to left-hand side
$2^n-1$2n−1	$=$=	$1023$1023	Divide both sides by $5$5
$2^n$2n	$=$=	$1024$1024	Add $1$1 to both sides
$\therefore n$∴n	$=$=	$10$10	Solve for $n$n, using guess and check, technology or logarithms.

Practice questions

Question 1

Question 2

Question 3

Question 4