5A, 5B - Finite geometric series

Repeated multiplication can make a long total surprisingly compact. In 5B, you will add finite geometric series and recover their parameters from sums. In 5A, you will use sigma bounds to identify the first included term and the number of terms being added.

The finite sum

How can you add $5+10+20+40+80+160$ without adding each new term to a running total?

Call the total $S$ . Multiplying by $2$ gives $2S=10+20+40+80+160+320$ . Subtract the original sum: all the middle terms cancel, leaving $2S-S=320-5$ , so $S=315$ .

The same cancellation works for a general finite geometric sum. In $S_n-rS_n$ , the terms $ar$ through $ar^{n-1}$ cancel, leaving $a-ar^n$ . This explains the formula; a formal proof is not needed.

Finite geometric sum

For a positive integer $n$ and $r\ne1$ ,

S_n=\frac{a(1-r^n)}{1-r}

Equivalently, $S_n=a(r^n-1)/(r-1)$ . Keep the numerator and denominator signs consistent.

On the formulae sheet.

There is no restriction that the ratio must be between $-1$ and $1$ for a finite sum. If $r=1$ , every term equals $a$ , so use $S_n=na$ . Substituting $r=1$ into the fraction would give an undefined zero denominator. For example, five terms each equal to $4$ total $20$ .

Worked example 1: an increasing finite sum

Question. Find the sum of the first six terms of a geometric series with $a=5$ and $r=2$ .

Plan. Substitute into the finite sum formula, retaining both negative signs.

\begin{aligned} S_6&=\frac{5(1-2^6)}{1-2} &&\text{substitute }n=6\\ &=\frac{5(-63)}{-1} &&\text{evaluate the power}\\ &=315 &&\text{divide} \end{aligned}

Answer. $315$ .

Check. Direct addition gives $5+10+20+40+80+160=315$ . All terms are positive, as is the answer.

Change only the sign of $r$ . The terms now alternate, and the last term can make the total negative.

Worked example 2: an alternating finite sum

Question. Find the sum of the first six terms when $a=5$ and $r=-2$ .

Plan. Put the entire negative ratio inside brackets before raising it to a power.

\begin{aligned} S_6&=\frac{5[1-(-2)^6]}{1-(-2)} &&\text{use the signed ratio}\\ &=\frac{5(1-64)}3 &&\text{even power is positive}\\ &=-105 &&\text{simplify} \end{aligned}

Answer. $-105$ .

Check. The first five terms total $5-10+20-40+80=55$ . The sixth is $-160$ , giving $55-160=-105$ .

In the opening total, five doublings produced six terms starting with $5$ , whose sum was $315$ . A finite total exists for either sign of ratio; convergence becomes relevant only when the number of terms is allowed to grow without bound.

Recovering a sequence from sums

When a problem gives sums instead of individual terms, write down exactly which terms each sum contains.

The first two terms total $S_2=a(1+r)$ . The next two total $ar^2+ar^3=r^2a(1+r)=r^2S_2$ . Therefore $S_4=S_2+r^2S_2$ . Grouping equal-length blocks can reveal the ratio more quickly than expanding two sum formulae.

Worked example 3: a positive first term

Question. A geometric series has $S_2=18$ , $S_4=90$ and a positive first term. Find $a$ and $r$ .

Plan. Compare the second block of two terms with the first, then use the sign of $a$ .

\begin{aligned} S_4-S_2&=r^2S_2 &&\text{third and fourth terms}\\ 90-18&=18r^2 &&\text{substitute the sums}\\ r^2&=4 &&\text{divide by }18\\ r&=2\quad\text{or}\quad r=-2 &&\text{retain both roots} \end{aligned}

Use $a(1+r)=18$ . If $r=2$ , then $a=6$ . If $r=-2$ , then $a=-18$ , which contradicts the positive first term.

Answer. $a=6$ and $r=2$ .

Check. The sequence begins $6,12,24,48$ . Its first two terms total $18$ , and all four total $90$ .

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5A, 5B - Finite geometric series

The finite sum

Recovering a sequence from sums

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