Finite Geometric Series | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Multiply, align, then subtract

A finite series adds a specified number of terms. Multiplying a GP sum by $r$ gives the same middle terms, plus one new term at the end.

S₄	rS₄	Subtract
a	0	a
ar	ar	0
ar²	ar²	0
ar³	ar³	0
0	ar⁴	−ar⁴

S₄ − rS₄ = a − ar⁴

The four-term example shows exactly which terms cancel.

For $n$ terms, subtracting $rS_n$ from $S_n$ leaves $a-ar^n$ . Factor and divide by $1-r$ :

(1-r)S_n=a(1-r^n)

S_n=\frac{a(1-r^n)}{1-r},\qquad r\ne1

MF19 supplies this formula. Multiplying its top and bottom by −1 gives the equally valid form $a(r^n-1)/(r-1)$ , often simpler for $r>1$ .

If $r=1$ , do not divide by zero. Add $a$ exactly $n$ times: $S_n=na$ . Finite sums do not require $|r|<1$ .

Use the number of terms in the power

Worked example

Find the sum of the first 8 terms of $3,6,12,\ldots$ .

S_8=\frac{3(2^8-1)}{2-1}=765

The last term uses $2^7$ , but the sum uses $2^8$ . Multiplying the row by 2 created that extra power.

For negative ratios, power the entire signed number. For $8,-4,2,-1$ :

S_4=\frac{8[1-(-1/2)^4]}{1-(-1/2)}=5

Check by direct addition. For a minimum count, use a calculator table of the sum at integer $n$ . For $3,6,12,\ldots$ , $S_8=765$ and $S_9=1533$ , so 9 terms first exceed 1000. Positive terms make these sums increase.

Now you try

Your turn9709/13 May/Jun 2024 Q10(a)–(b) [5 marks]

A GP $a_1,a_2,a_3,\ldots$ has first term 2 and positive ratio $r$ . Given $\tfrac92a_5+7a_3=8$ , find $r$ .

Then find the sum of its first 20 terms to 4 significant figures.

Show worked answer

$a_3,a_5$ mean term positions, just like $u_3,u_5$ . Substitute $a_3=2r^2$ and $a_5=2r^4$ :

9r^4+14r^2-8=0

Set $y=r^2$ . The polynomial solver gives $y=4/9,-2$ ; the convenient roots give

(9y-4)(y+2)=0

Reject $y=-2$ because a real square cannot be negative. $r=\pm2/3$ , and the given positive ratio selects $r=2/3$ .

S_{20}=\frac{2[1-(2/3)^{20}]}{1-2/3}

Answer

r=\frac23,\qquad S_{20}=5.998