In the summation:

$\sum\limits_{j=2}^n (j-1) = \frac{n(n-1)}{2}$

Given that $\sum\limits_{j=2}^n (j) = \frac{n(n+1)}{2}-1$. Expanding it: $ \frac{n(n+1)-2n(n+1)}{2}$ = $ \frac{-n(n+1)}{2}$ and bringing the minus sign inside $ \frac{n(n-1)}{2}$. So that would mean $\sum\limits_{j=2}^n (j) = \frac{n(n-1)}{2}$ but that's the result of $\sum\limits_{j=2}^n (j-1)$.
What I'm doing wrong?



$\sum_{j=2}^{n} (j-1) = \sum_{j=2}^{n} j - \sum_{j=2}^{n} 1= \frac{n(n+1)}{2} - n = \frac{n^2+n-2n}{2} = \frac{n^2-n}{2} = \frac{n(n-1)}{2}$. (Note that the 1's cancel out)

As for your arithmetic:

$1) \frac{n(n+1)}{2} - n = \frac{n(n+1)-2n}{2} \neq \frac{n(n+1)-2n(n+1)}{2}; \\ 2) \frac{-n(n+1)}{2} = \frac{n(-n-1)}{2} \neq \frac{n(n-1)}{2}.$

Because you've erred twice, your conclusion is nonsensical.

  • $\begingroup$ @user112780: does this help? $\endgroup$ – Chris K Dec 5 '13 at 4:35
  • $\begingroup$ Yes it helps a lot! I got it! Thanks a lot for correcting my math. Now I understand. $\endgroup$ – user112780 Dec 5 '13 at 8:36

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