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sum of the first m terms of n^3/2^3 = ?


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#1 zerocordas

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Posted 21 September 2016 - 10:24 AM

from a high school level math book "les math au carré" by M-P. Falissard , page 133

 

n integer

 

she uses the expression to deduce the limit of the sum as m approaches infinity (26)

 

 

my gratitude to any solid help on this one !


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#2 HallsofIvy

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Posted 21 September 2016 - 12:23 PM

First, of course, you can factor the "2^3 = 8" out of the series: \sum_{n=1}^m\frac{n^3}{2^3}= \frac{1}{8}\sum_{n=1}^m n^3.

 

 Second, the sum of n to a power is always a polynomial in n of degree one higher than that power.

 

 So \sum_{n=1}^m n^3= am^4+ bm^3+ cm^2+ dm+ e.  Check the value for 4 different values of m to get 4 equations for a, b, c, d, and e.

 

  For example, if m= 1, \sum_{n= 1}^1 n^3= 1^3= 1 so a+ b+ c+ d+ e= 1.  If m= 2, \sum_{n=1}^2= 1^3+ 2^3= 1+ 8= 9 so 16a+ 8b+ 4c+ 2d+ e= 9.

 

  Now look at m= 3 and 4 to get two more equations.


Edited by HallsofIvy, 21 September 2016 - 12:27 PM.

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#3 zerocordas

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Posted 21 September 2016 - 02:24 PM

sorry for an unforgivable typo !!!!

 

n^3/2^n .

 

but even this answer was helpful. sum of the powers must be the next power , seems logical , is there a proof ?

 

fantastic ,    (n(n+1))^2   /   4     is a polynome of deg 4 indeed.


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