0


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?

user112780
  • 317
  • 1
  • 3
  • 6

1 Answers1

3

So...

$\sum_{j=2}^{n} (j-1) = \sum_{j=2}^{n} j - \sum_{j=2}^{n} 1= \frac{n(n+1)}{2} - n = \frac{n^2+n-2n}{2} = \frac{n^2-n}{2} = \frac{n(n-1)}{2}$. (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.

Christopher K
  • 3,383