Sequences and Series

Hong Kong

Stage 4 - Stage 5

Lesson

If the first term of a geometric progression is $t_1=a$`t`1=`a` and the common ratio is $r$`r`, then the sequence becomes:

$t_1=a,t_2=ar,t_3=ar^2,t_4=ar^3,t_5=ar^4,\ldots$`t`1=`a`,`t`2=`a``r`,`t`3=`a``r`2,`t`4=`a``r`3,`t`5=`a``r`4,…

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:

$S_n=a+ar+ar^2+ar^3+...+ar^{n-1}$`S``n`=`a`+`a``r`+`a``r`2+`a``r`3+...+`a``r``n`−1

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

$rS_n=ar+ar^2+ar^3+...+ar^{n-1}+ar^n$`r``S``n`=`a``r`+`a``r`2+`a``r`3+...+`a``r``n`−1+`a``r``n`

Then, by carefully subtracting $rS_n$`r``S``n` from $S_n$`S``n` term by term, we see that **all of the middle terms disappear**:

$S_n-rS_n=a+\left(ar-ar\right)+\left(ar^2-ar^2\right)+...+\left(ar^{n-1}-ar^{n-1}\right)-ar^n$`S``n`−`r``S``n`=`a`+(`a``r`−`a``r`)+(`a``r`2−`a``r`2)+...+(`a``r``n`−1−`a``r``n`−1)−`a``r``n`

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

$S_n=\frac{a\left(1-r^n\right)}{1-r}$`S``n`=`a`(1−`r``n`)1−`r`

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

$S_n=\frac{a\left(r^n-1\right)}{r-1}$`S``n`=`a`(`r``n`−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:

$S_{10}=\frac{96\times\left(1-\left(\frac{1}{2}\right)^{10}\right)}{1-\left(\frac{1}{2}\right)}=191.8125$`S`10=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:

$S_{10}=\frac{2\times\left(3^{20}-1\right)}{3-1}=3486784400$`S`10=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:

$S_n=a+a+a+...+a=a\times n$`S``n`=`a`+`a`+`a`+...+`a`=`a`×`n`

Consider the series $5+10+20$5+10+20 ...

Find the sum of the first $12$12 terms.

Consider the number $0.252525$0.252525$\ldots$…

Which option represents it as an infinite geometric series?

$0.25+0.0025+0.000025+\ldots$0.25+0.0025+0.000025+…

A$0.25+0.025+0.000025+\ldots$0.25+0.025+0.000025+…

B$0.25+0.0025+0.00025+\ldots$0.25+0.0025+0.00025+…

CTherefore express $0.252525$0.252525$\ldots$… as a fraction.