How do I write an equation to find the sum of this series?
January 21st, 2014 at 10:14:58 AM
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I want to find the sum of a series like this:
n+(n*.99)+(n*.99^2)+(n+.99^3)...
Say I want to find the sum of that series for x iterations. How do I do that?
n+(n*.99)+(n*.99^2)+(n+.99^3)...
Say I want to find the sum of that series for x iterations. How do I do that?
January 21st, 2014 at 10:30:05 AM
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Use the sigma sign? Also, I'd change your n to an x.
sigma (1 to n) x+ (x*.99^n)
sigma (1 to n) x+ (x*.99^n)
January 21st, 2014 at 10:40:50 AM
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check out:
Summing a Geometric Series
http://www.mathsisfun.com/algebra/sequences-sums-geometric.html
Use this formula:
a(1-r^n/1-r)
a is the first term
r is the "common ratio" between terms
n is the number of terms
they even show examples that I can grasp
Summing a Geometric Series
http://www.mathsisfun.com/algebra/sequences-sums-geometric.html
Use this formula:
a(1-r^n/1-r)
a is the first term
r is the "common ratio" between terms
n is the number of terms
they even show examples that I can grasp
winsome johnny (not Win some johnny)
January 21st, 2014 at 1:11:21 PM
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Quote: TheJacob
n+(n*.99)+(n*.99^2)+(n+.99^3)...
= n * (.99^0 + .99 + .99^2 + .99^3 + ... + .99^x)
= n * (1 - .99^(x+1)) / (1 - .99)
See for "geometric series".
January 21st, 2014 at 3:46:47 PM
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See lines 6 and 7 on http://mathworld.wolfram.com/GeometricSeries.html
Essentially it's easier when considering your question to ignore the n, as that is a common factor, and just consider the sum of 1+x+x^2+x^3... where 0<x<1. This infinite sum is 1/(1-x). (for instance 1 + 1/2 + 1/4 ... = 1/(1-1/2) = 2)
Simple proof
S = 1+x+x^2....
S * x = x+x^2+x^3....
S - S*x = S(1-x) = (1+x+x^2...)-(x+x^2...) = 1
S = 1 / (1-x)
A finite sum (the first N terms) can be considered from two series (1st to infinity) and (N+1th to infinity) and taking the difference.
(i) The infinite sum 1+x+..... = 1/(1-x).
(ii) The infinite sum starting just after the Nth term (x^(N+1)+x^(N+2)...) which is x^(N+1) * the infinite sum in (i). Thus it's value is x^(N+1)/(1-x).
The difference gives the answer which is therefore (1-x^(N+1))/(1-x).
Essentially it's easier when considering your question to ignore the n, as that is a common factor, and just consider the sum of 1+x+x^2+x^3... where 0<x<1. This infinite sum is 1/(1-x). (for instance 1 + 1/2 + 1/4 ... = 1/(1-1/2) = 2)
Simple proof
S = 1+x+x^2....
S * x = x+x^2+x^3....
S - S*x = S(1-x) = (1+x+x^2...)-(x+x^2...) = 1
S = 1 / (1-x)
A finite sum (the first N terms) can be considered from two series (1st to infinity) and (N+1th to infinity) and taking the difference.
(i) The infinite sum 1+x+..... = 1/(1-x).
(ii) The infinite sum starting just after the Nth term (x^(N+1)+x^(N+2)...) which is x^(N+1) * the infinite sum in (i). Thus it's value is x^(N+1)/(1-x).
The difference gives the answer which is therefore (1-x^(N+1))/(1-x).
