Geometric Series

If the first term of a geometric progression is $t_1=a$t1​=a and the common ratio is $r$r, then the sequence becomes:

$t_1=a,t_2=ar,t_3=ar^2,t_4=ar^3,t_5=ar^4,\ldots$t1​=a,t2​=ar,t3​=ar2,t4​=ar3,t5​=ar4,…

Suppose we wish to add the first $n$n terms of this sequence. Technically, this is known as a series, or alternatively the partial sum of the sequence. We could write the sum as:

$S_n=a+ar+ar^2+ar^3+...+ar^{n-1}$Sn​=a+ar+ar2+ar3+...+arn−1

If we multiply both sides of this equation by the common ratio $r$r we see that:

$rS_n=ar+ar^2+ar^3+...+ar^{n-1}+ar^n$rSn​=ar+ar2+ar3+...+arn−1+arn

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

$S_n-rS_n=a+\left(ar-ar\right)+\left(ar^2-ar^2\right)+...+\left(ar^{n-1}-ar^{n-1}\right)-ar^n$Sn​−rSn​=a+(ar−ar)+(ar2−ar2)+...+(arn−1−arn−1)−arn

This means that $S_n-rS_n=a-ar^n$Sn​−rSn​=a−arn and when common factors are taken out on both sides of this equation, we find $S_n\left(1-r\right)=a\left(1-r^n\right)$Sn​(1−r)=a(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:

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

An extra step, multiplying the numerator and denominator by $-1$−1, reveals a slightly different form for $S_n$Sn​:

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

This form is usually easier to use when $\left|r\right|>1$|r|>1.

 

As an example, adding up the first $10$10 terms of the geometric progression that begins $96+48+12+...$96+48+12+... is straightforward. Since $a=96$a=96 and $r=\frac{1}{2}$r=12​ we have:

$S_{10}=\frac{96\times\left(1-\left(\frac{1}{2}\right)^{10}\right)}{1-\left(\frac{1}{2}\right)}=191.8125$S10​=96×(1−(12​)10)1−(12​)​=191.8125

To add up the first $20$20 terms of the sequence $2,6,18,...$2,6,18,... we might use the second version of the formula to reveal:

$S_{10}=\frac{2\times\left(3^{20}-1\right)}{3-1}=3486784400$S10​=2×(320−1)3−1​=3486784400

 

You might be wondering that the two forms of the sum formula excludes the case for $r=1$r=1. This is not an issue, for if $r=1$r=1, then the series becomes:

$S_n=a+a+a+...+a=a\times n$Sn​=a+a+a+...+a=a×n

 

Practice questions

Question 1

QUESTION 2

QUESTION 3