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Write down the sum of $\displaystyle \sum_1^{2N} n^3$ in terms of $N$, and hence find:
$1^3 - 2^3 + 3^3 - 4^3 + \cdots - (2N)^3$ in terms of $N$, simplifying your answer.

I found this to be $n^2(2n+1)^2$ but the next part is not making sense to me.

Why is the general term of this sum $-(2N)^3$, where it doesn't work for N=0 etc?

thanks

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  • $\begingroup$ what do you mean? The general term of this sum is not $-(2N)^2$. And what do you mean by "where it doesn't work for $N=0$ etc"? $\endgroup$ – Gerry Myerson Aug 25 '13 at 13:33
  • $\begingroup$ I thought that the last term in the sum was the general term? like Sum = 1 + 2 + 3 + .... + n then the general term was n? $\endgroup$ – salman Aug 25 '13 at 13:35
  • $\begingroup$ The last term in the series is not $-(2N)^2$, either. $\endgroup$ – Gerry Myerson Aug 25 '13 at 13:36
  • $\begingroup$ In a finite series like this, what is meant by the $-(2N)^3$ $\endgroup$ – salman Aug 25 '13 at 13:46
  • $\begingroup$ It's the last term in the sum. $\endgroup$ – Gerry Myerson Aug 25 '13 at 13:50
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HINT:

As $$ \begin{align} & \sum_{1\le r\le n}r^3=\frac{n^2(n+1)^2}4 \\[10pt] & 1^3 - 2^3 + 3^3 - 4^3 + \cdots - (2N)^3 \\[10pt] & =\sum_{1\le r\le 2N}r^3-2\sum_{1\le r\le N}(2r)^3 \\[10pt] & =\sum_{1\le r\le 2N}r^3-2\cdot8\sum_{1\le r\le N}r^3 \\[10pt] & =\frac{(2N)^2(2N+1)^2}4-16\frac{N^2(N+1)^2}4 \\[10pt] & =N^2\{(2N+1)^2-4(N+1)^2\} \\[10pt] & =-N^2(4N+3) \end{align} $$

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  • $\begingroup$ @MichaelHardy, could you please add displaystyle in the fourth line and replace brace with bracket in the last line else my change will add new edition $\endgroup$ – lab bhattacharjee Aug 25 '13 at 13:31
  • $\begingroup$ This formula gives the wrong answer for $N=1$. Also, I don't think it engages with the question OP is asking (which is understandable, since the question is not). $\endgroup$ – Gerry Myerson Aug 25 '13 at 13:32
  • $\begingroup$ How do I write the sum as a summation? what's the general term of the sequence $1^3 -2^3+3^3-4^3+..-(2N)^3$ $\endgroup$ – salman Aug 25 '13 at 13:39
  • $\begingroup$ @user90771, have you noticed the latest edition of this answer. The general term is $(-r)^3=-r^3,1\le r\le 2N$ $\endgroup$ – lab bhattacharjee Aug 25 '13 at 13:40
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    $\begingroup$ @user90771, observe that the sign toggles, hence $(-1)^{r-1}$ as the sign is negative for the even terms and positive for the odd terms $\endgroup$ – lab bhattacharjee Aug 25 '13 at 13:50

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