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 sum Up (Posted on 2014-02-26)
Find the infinite sum of :: 13/31+23/32+33/33+43/34+...

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 re(2): Method for solution... Comment 5 of 5 |
(In reply to re: Method for solution... by Jer)

I guess this is the method that tomarken referred to:

1 + x + x2 + x3 + x4 + ..             = 1/(1 – x)

Alternately, differentiate wrt x and multiply by x to regain the powers:

1 + 2x + 3x2 + 4x3 +..                = (1–x)-2

x + 2x2 + 3x3 + 4x4 +..               = x(1–x)-2

1 + 22x + 32x2 + 42x3 +.. = 2x(1–x)-3 + (1–x)-2

x + 22x2 + 32x3 + 42x4 +..            = 2x2(1–x)-3 + x(1–x)-2

1 + 23x + 33x2 + 43x3 +.. = 6x2(1–x)-4 + 4x(1–x)-3 + 2x(1–x)-3 + (1–x)-2

x + 23x2 + 33x3 + 43x4 +.. = 6x3(1-x)-4 + 6x2(1-x)-3 + x(1-x)-2

Now substituting x = 1/3 gives:

13/31 + 23/32 + 33/33 + 43/34 + ...   =   33/8

 Posted by Harry on 2014-03-06 12:55:19

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