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zerocordas

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

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zerocordas    0

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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HallsofIvy    52

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

 

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

 

So [math]\sum_{n=1}^m n^3= am^4+ bm^3+ cm^2+ dm+ e[/math]. 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, [math]\sum_{n= 1}^1 n^3= 1^3= 1[/math] so a+ b+ c+ d+ e= 1. If m= 2, [math]\sum_{n=1}^2= 1^3+ 2^3= 1+ 8= 9[/math] so 16a+ 8b+ 4c+ 2d+ e= 9.

 

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

Edited by HallsofIvy

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zerocordas    0

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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