zerocordas Posted September 21, 2016 Share Posted September 21, 2016 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 ! Link to comment Share on other sites More sharing options...
Country Boy Posted September 21, 2016 Share Posted September 21, 2016 (edited) 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 September 21, 2016 by Country Boy Link to comment Share on other sites More sharing options...
zerocordas Posted September 21, 2016 Author Share Posted September 21, 2016 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. Link to comment Share on other sites More sharing options...
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