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

## 3 posts in this topic

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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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
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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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