# Summation of stopping point

I'm learning summation and I need help with the following sum:

$\sum\limits_{i=0}^{n-2} n$

What I thoughts is, since $n$ don't change, my sum will be

$S_n$ = $n_1$ + $n_2$ + $n_3$ + ... + $n_{n-2}$

Then:

$2S_n$ = ($n_1$+$n_{n-2}$)$n$

$2S_n$ = ($2n$)$n$ = $2n^2$

$S_n$ = $n^2$

But the answer in my book is $n(n-1)$ and I dont get why.

Thanks.

Since $n$ doesn't change as $i$ varies from $0$ to $n-2$, you're summing multiple copies of $n$. How many copies? There are $n-1$ items in your sum, since the starting point $i=0$ contributes an item beyond the stopping point $i=n-2$, and $(n-2)+1 = n-1$. This makes a total of $n(n-1)$.
Since $n$ is a constant $$\sum\limits_{i=0}^{n-2} n=n\sum\limits_{i=0}^{n-2} 1$$ Now, just count.