I need to find the sum of this series:
$1^3$, $-2^3$, $3^3$, $-4^3$, ... ,$2n^3$
so I've split that up into:
$\sum_{r = 1}^n (2r-1)^3 - \sum_{r = 1}^n (2r)^3$
So all I need is how to find: $\sum_{r = 1}^n (2r-1)^3$
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$(2r-1)^3=8r^3-12r^2+6r-1$. – vadim123 Oct 05 '15 at 18:48
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It seems, from the first terms, that the general term of the sequence is $$a_n=(-1)^{n+1}n^3$$ Is this right? – Ángel Mario Gallegos Oct 05 '15 at 18:49
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Yeah that is the general term – Chris Fullerton Oct 05 '15 at 18:51
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$\sum\limits_{r=1}^{n}(2r-1)^3 - \sum\limits_{r=1}^{n}(2r)^3 = \sum\limits_{r=1}^{n}8r^3 - \sum\limits_{r=1}^{n}12r^2+\sum\limits_{r=1}^{n}6r - \sum\limits_{r=1}^{n}1 - \sum\limits_{r=1}^{n}8r^3 = -12\sum\limits_{r=1}^{n}r^2 + 6\sum\limits_{r=1}^{n}r - \sum\limits_{r=1}^{n}1 = -2n(n+1)(2n+1)+3n(n+1)-n = -4n^3-3n^2$ – barak manos Oct 05 '15 at 18:54
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What are you looking for $$\displaystyle{\sum_{k=1}^n(-1)^{k+1}k^3\quad\text{or}\quad\sum_{k=1}^n[(2k-1)^3-(2k)^3]};;;{\Huge \color{red}{?}}$$ – Ángel Mario Gallegos Oct 05 '15 at 19:01
2 Answers
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HINT: notice $$\sum_{r=1}^{n}(2r-1)^3$$ $$=\sum_{r=1}^{n}\left((2r)^2-(1)^3-3(2r)(1)(2r-1)\right)$$ $$=\sum_{r=1}^{n}\left(8r^3-12r^2+6r-1\right)$$ $$=8\sum_{r=1}^{n}r^3-12\sum_{r=1}^{n}r^2+6\sum_{r=1}^{n}r-\sum_{r=1}^{n}1$$ $$=8\left(\frac{n(n+1)}{2}\right)^2-12\left(\frac{n(n+1)(2n+1)}{6}\right)+6\left(\frac{n(n+1)}{2}\right)-n$$ You can solve further by simplifying
Harish Chandra Rajpoot
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Hint : The required(original one i.e. $1^3 - 2^3 + 3^3 - ... - (2n)^3$) sum is $\sum_{r = 1}^{2n} r^3 - 16\sum_{r = 1}^n r^3.$ Use the formula to calculate the sum of $1^{st} n$ cubes.
Ojas
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You should be clear you are talking about the initial sum, not the sum derived later. – Thomas Andrews Oct 05 '15 at 19:04