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?


  • $\begingroup$ 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


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

  • $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.