2021-11-26, 14:13 | #1 |
"Matthew Anderson"
Dec 2010
Oregon, USA
10001001101_{2} Posts |
geometric series
Many of you are familiar with geometric series. Here is a little derivation of a common result.
Finite Geometric Series Let S1 = 1 + a + a^2 + ... + a^n. We multiply S1 by ‘a’ then see a*S1 = a+ a^2 + … + a^(n+1). Subtract the second equation from the first one. (1-a)*S1 = 1-a^(n+1). Therefore S1 = [1-a^(n+1)]/(1-a). We are sure of this. This result about finite geometric series is in many textbooks. The Wikipedia on this is very good. The infinite case is another story. If S2 = 1 + b + b^2 + … is an infinite sum then S2 converges for -1<b<1. |
Thread Tools | |
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Prime Numbers as a Function of a Geometric Progression | danupp | And now for something completely different | 2 | 2017-11-04 17:55 |
Best 4XX series GPU | siegert81 | GPU Computing | 47 | 2011-10-14 00:49 |
Geometric Combinatorial Puzzle | davar55 | Puzzles | 14 | 2006-04-26 17:27 |
Another Series | Gary Edstrom | Puzzles | 7 | 2003-07-03 08:32 |
Series | Rosenfeld | Puzzles | 2 | 2003-07-01 17:41 |