What is the value of $\displaystyle \sum_{n=0}^{\infty} \sum_{j=0}^{n} \sum_{k=0}^{n} \binom{n} {k} \binom{n} {j} (-1)^{j+k}$?

Unfortunately and to be honest, I have no idea to solve this problem. One might be progressed, but this is triple summation, you see. I have looked back at my textbook about this section, but found nothing about triple summation. Could you help me, at least give something brilliant to clear the trouble?

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    $\begingroup$ I think you can rewrite it this way: $\sum\limits_{n=0}^{\infty}\left(\left(\sum\limits_{k=0}^{n} \binom{n} {k}(-1)^k \right)\left(\sum\limits_{j=0}^{n} \binom{n} {j}(-1)^j \right)\right)=\sum\limits_{n=0}^{\infty} \left(\sum\limits_{i=0}^n \binom{n}{i} (-1)^i\right)^2$ $\endgroup$ – Botond Feb 14 '18 at 17:23

Notice that by expanding $(1-1)^n$ we obtain $ \sum_{k=0}^{n} \binom{n} {k} (-1)^{k}$. Moreover $$\sum_{j=0}^{n} \sum_{k=0}^{n} \binom{n} {k} \binom{n} {j} (-1)^{j+k}= \sum_{j=0}^{n}\binom{n} {j} (-1)^{j} \sum_{k=0}^{n} \binom{n} {k} (-1)^{k}.$$ Can you take it from here?

  • $\begingroup$ The answer is 0? $\endgroup$ – Shane Dizzy Sukardy Feb 15 '18 at 3:02
  • $\begingroup$ Yes, it is. This triple sum is not so scary after all ;-) $\endgroup$ – Robert Z Feb 15 '18 at 6:59

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