Why the arithmetic trick fails here.
arithmetic
2, 5, 8, 11 …
ends pair to a constant
geometric
3, 6, 12, 24 …
3 + 384 ≠ 6 + 192
387 versus 198 — the pairs are nothing alike, so a different formula is needed.
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Finite Geometric Series
Adding a doubling sequence — the terms grow, so pairing will not help.
The problem
Find the sum of the first 8 terms of 3, 6, 12, 24, …
S = a(rⁿ − 1)r − 1 = 3(2⁸ − 1)2 − 1 = 3(256 − 1) = 3 × 255 = 765. D forgets the −1 inside the bracket, which is the difference between the sum and one term too many.
Read off a, r and n.
first term a
3
ratio r
2
terms n
8
Sₙ = a(1 − rⁿ)1 − r
Divide any term by the one before to get r: 6 ÷ 3 = 2, 12 ÷ 6 = 2. Confirm on more than one pair.
Substitute carefully.
S₈ = 3(1 − 2⁸) ÷ (1 − 2)
drop the values in
2⁸ = 256
the exponent first
= 3(−255) ÷ (−1)
both come out negative
= 765
the negatives cancel
Two negatives, one answer
With r > 1 both the bracket and the denominator go negative and cancel. Losing one of the signs turns a sum of 765 into −765.
When the terms shrink.
16 + 8 + 4 + 2, so a = 16, r = ½, n = 4.
S₄ = 16(1 − (½)⁴) ÷ (1 − ½)
substitute
(½)⁴ = 116
the exponent
= 16(1516) ÷ (½)
tidy the bracket
= 15 ÷ 0.5 = 30
and 16+8+4+2 = 30 ✓
With |r| < 1 everything stays positive, and no sign juggling is needed.
Where the infinite formula comes from.
if |r| < 1, rⁿ shrinks towards 0
as n grows
S = a(1 − 0) ÷ (1 − r)
the term disappears
S = a ÷ (1 − r)
the infinite sum
So 16 + 8 + 4 + … forever totals 16 ÷ 0.5 = 32. Our four terms reached 30, and the rest of eternity adds only 2.
Only when |r| < 1
With r = 2 the terms keep growing, and 2ⁿ never settles. An infinite sum simply does not exist there.
A negative ratio: keep it in brackets.
a = 4, r = −12 ÷ 4 = −3
the ratio carries the sign
S₄ = 4(1 − (−3)4)1 − (−3)
brackets round −3
= 4(1 − 81)4
(−3)4 = +81, and 1 + 3 = 4
= −80
Check: 4 − 12 + 36 − 108 = −80 ✓.
−34 is not (−3)4
Dropping the brackets gives −81 instead of +81, and the sum comes out as 82. An even power of a negative ratio is positive.
Backwards: find n from the sum.
Sₙ = 3(4n − 1)4 − 1 = 4n − 1
for r > 1 use a(rⁿ − 1) ÷ (r − 1)
4n − 1 = 1023
set it equal to the sum
4n = 1024 = 45
a power of the ratio
n = 5
match the exponents
Check: 3 + 12 + 48 + 192 + 768 = 1023 ✓.
Two sums given: the second half is the first half times rⁿ.
Terms 4 to 6 are terms 1 to 3, each multiplied by r³.
S₆ = S₃ + r³S₃ = S₃(1 + r³)
first half plus r³ times the first half
63 = 7(1 + r³), so 1 + r³ = 9
divide by 7
r³ = 8, r = 2
cube root
a(1 + 2 + 4) = 7, so a = 1
back into S₃
Check: 1 + 2 + 4 = 7 and 8 + 16 + 32 = 56; 7 + 56 = 63 ✓.
Variations: a negative fraction r, the last term instead of n, r = 1.
Negative fraction r. Find 32 − 16 + 8 − 4 + 2.
a = 32, r = −12, n = 5
−16 ÷ 32 = −12
S₅ = 32(1 − (−12)5) ÷ (1 + 12)
brackets round −12
= 32(3332) ÷ (32) = 33 × 23
(−12)5 = −132
= 22
and 32 − 16 + 8 − 4 + 2 = 22 ✓
What is different: two signs to watch at once. The odd power keeps (−12)5 negative, so 1 − (−132) becomes 1 + 132.
The last term instead of n. Find 2 + 6 + 18 + … + 4374.
S = (r · last − a) ÷ (r − 1)
rⁿ · a is one step past the last term
S = (3 × 4374 − 2) ÷ (3 − 1)
a = 2, r = 3
S = 13 120 ÷ 2 = 6560
no need to find n
What is different: n is never needed. Check the long way: 4374 = 2 × 37, so n = 8 and 2(38 − 1) ÷ 2 = 6560 ✓.
r = 1. Find the sum of the first 12 terms of 5 + 5 + 5 + …
r = 5 ÷ 5 = 1
every term is the same
a(1 − 112) ÷ (1 − 1) = 0 ÷ 0
the formula breaks
S₁₂ = 12 × 5 = 60
just count the copies
What is different: the formula divides by 1 − r, so it cannot be used when r = 1. The sum is n × a.
The bouncing ball: falls and rises separately.
Every rise is followed by a fall, and the first drop has no rise before it — so there is one rise fewer than falls.
falls: 32 + 16 + 8 + 4 = 60
four falls
rises: 16 + 8 + 4 = 28
three rises before the 4th landing
total = 60 + 28 = 88 m
Shortcut: twice the falls, minus the first drop: 2(60) − 32 = 88 ✓.
One rise too many
Adding a fourth rise gives 90 m, and doubling all the falls gives 120 m. Both count bounces that have not happened by the 4th landing.
On the exam
Expect the sum to be given and n wanted, two sums such as S₃ and S₆, a negative ratio, or a story: chain messages, savings that double, a bouncing ball. With r > 1 write Sₙ = a(rⁿ − 1) ÷ (r − 1) and the signs look after themselves. Ten-second check: with five or six terms, add them by hand.
Check: add the eight terms.
the terms
3, 6, 12, 24, 48, 96, 192, 384
doubling each time
first four
3+6+12+24 = 45
—
last four
48+96+192+384 = 720
—
total
45 + 720 = 765
✓
Notice the sum is just under twice the last term. That is always true when r = 2.
Now you try
Practice problem
The problem
Find the sum of the first 6 terms of 5, 15, 45, …
S = 5(3⁶ − 1)3 − 1 = 5(729 − 1)2 = 5 × 728 ÷ 2 = 1,820. A forgets to divide by r − 1 = 2. When r is not 2, that denominator matters a great deal.