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Evaluate $$\sum_{k = -3}^2 k^3$$

If I evaluate it using the expression above by putting the lower and upper limits it produces $-27$ for an answer, but when evaluated using the closed form expression of the geometric sequence where I put first term $a = -27$, $n = 3$ and common ratio $r = 3$, it produces the different result.

I believe my common ratio $r = 3$ is incorrect. Can anyone tell me is $r$ value wrong? If yes, then how to calculate common ratio involving the cube or square etc.

EDIT: Earlier I thought it was geometric sequence, but it is arithmetic sequence so I'm looking for d (common difference).

EDIT I: As @Brian M. Scott pointed out, it is neither an arithmetic sequence nor geometric sequence.

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    $\begingroup$ It isn’t a geometric sequence: there is no constant $r$ such that $(k+1)^3=rk^3$ for each $k$. $\endgroup$
    – Brian M. Scott
    Oct 22, 2016 at 18:35
  • $\begingroup$ @BrianM.Scott, yeah you are right. I think then it is an arithmetic sequence. If yes, then what is d? (common difference) Thanks a lot. $\endgroup$
    – Anonymous
    Oct 22, 2016 at 18:40
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    $\begingroup$ It isn’t an arithmetic sequence either, so you need either brute force (which with only six terms isn’t too awful) or a bit of cleverness. The easiest way to evaluate it is to notice that $(-k)^3=-k^3$, so that the $k=-2$ and $k=2$ terms cancel, the $k=-1$ and $k=1$ terms cancel, the $k=0$ term contributes nothing, and all that actually matters is the $k=-3$ term, making the total $-27$. $\endgroup$
    – Brian M. Scott
    Oct 22, 2016 at 18:43

4 Answers 4

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\begin{align} S &= \sum_{k=-3}^2 k^3 \\ &= (-3)^3 + (-2)^3 + (-1)^3+ 0^3 + 1^3 + 2^3 \\ &= (-3)^3 - (2)^3 + 2^3 - 1 + 1 \\ &= -27 \end{align}

In general you could transform into a combination of sums of $j^k$, which is the topic of Faulhaber's formula. \begin{align} S &= \sum_{k=-3}^2 k^3 \\ &= \sum_{j=1}^6 (j-4)^3 \\ &= \sum_{j=1}^6 (j-4)(j^2 - 8j + 16) \\ &= \sum_{j=1}^6 (j^3 - 12j^2 + 48j-64) \\ &= \sum_{j=1}^6 j^3 - 12 \sum_{j=1}^6 j^2 + 48 \sum_{j=1}^6 j- 64 \sum_{j=1}^6 1 \\ &= \left. \left( \frac{n(n+1)}{2} \right)^2\right\vert_{n=6} -12 \left. \frac{n(n+1)(2n+1)}{6} \right\vert_{n=6} +48 \left. \frac{n(n+1)}{2}\right\vert_{n=6} - 64 \cdot 6 \\ &= 441 - 1092 + 1008 - 384 \\ &= -27 \end{align} which is overkill manually, but nice for a computer.

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You could be interessed in this.

You can split the sum in two parts : $$ \sum_{k=-3}^2 k^3 = \sum_{k=-3}^{-1} k^3 + \sum_{k=0}^2 k^3 = -\sum_{k=1}^3 k^3 + \sum_{k=0}^2 k^3 $$ and apply the "power of $3$" formula on both sums.

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Try to generalize this notion. Note that the sum of the numbers within a single color is a perfect cube.

enter image description here

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Summation of k^3 has a pretty simple formula, and a decent derivation. Check it out on Wolframalpha.

http://mathworld.wolfram.com/PowerSum.html

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