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

Question

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, Dec 05 '13 at 02:53
Of course. It was so obvious. I can't believe I couldn't see it. Thanks for the tip — user112780, Dec 05 '13 at 03:01

score 1 · Accepted Answer · answered Dec 05 '13 at 02:58

1

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

answered Dec 05 '13 at 02:58

ncmathsadist

49,383

Vey clear thank you. It was so obvious!!!! Thanks for taking time to pointing it out! I got it ! – user112780 Dec 05 '13 at 03:02

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

1 Answers1