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Geometric series and sum to infinity
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In a geometric sequence you multiply by the same number each time. If that number is small enough, the series adds up to a finite total even with infinitely many terms.
The nth term and sum
First term a, common ratio r.
un = arn − 1
Sn = a(1 − rn)(1 − r) (or a(rn − 1)(r − 1), easier when r > 1)
un = arn − 1
Sn = a(1 − rn)(1 − r) (or a(rn − 1)(r − 1), easier when r > 1)
Sum to infinity
If |r| < 1, the terms shrink towards 0 and the sum gets closer and closer to
S∞ = a(1 − r)
24 + 12 + 6 + ...: a = 24, r = 12, S∞ = 24(1/2) = 48
If |r| ≥ 1 there is no sum to infinity.
S∞ = a(1 − r)
24 + 12 + 6 + ...: a = 24, r = 12, S∞ = 24(1/2) = 48
If |r| ≥ 1 there is no sum to infinity.
Finding r
u2 = 6 and u5 = 162:
ar = 6 and ar4 = 162
Divide: r3 = 27, so r = 3
ar = 6 and ar4 = 162
Divide: r3 = 27, so r = 3
Find the sum of the first 8 terms of 3 + 6 + 12 + ...
- a = 3, r = 2
- S8 = 3(28 − 1)(2 − 1)
- 3 × 255
Answer: 765
un = arn − 1, Sn = a(1 − rn)(1 − r), and S∞ = a(1 − r) only when |r| < 1.
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