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

I've know that $\sum\limits_{i=1}^n k$ is $\frac{n(n+1)}{2}$.
Then why $\sum\limits_{j=2}^n j = \frac{n(n+1)}{2} -1$ ? I understand that i=1 $\neq$ j=2 is the key, but I can't get further.

from : Introduction to Algorithms (3rd edition) p. 27

• Note $\sum\limits_{j=1}^n j = \frac{n(n+1)}{2}$ and $1+\sum\limits_{j=2}^n j=\sum\limits_{j=1}^n j$. – Pedro Tamaroff Dec 5 '13 at 2:53
• Of course. It was so obvious. I can't believe I couldn't see it. Thanks for the tip – user112780 Dec 5 '13 at 3:01

$${n(n+1)\over 2}=\sum_{k=1}^n k = 1 + \sum_{k=2}^n k.$$