# How do we solve complicated summations

I am reading Introduction with Algorithms book and my first doubt arises in time analysis of my first insertions sort's algorithm uses these sigma problems. I learned about this in junior classes but never solved complicated ones.Even appendix in the book discuss very little about complicated sigma addition. Somebody answer the solution for two of the problems mentioned below?

I know:

$$\sum^n_{j=1}j = n(n+1)/2$$

but how these two evaluated?

$$\sum^n_{j=2}j = n(n+1)/2-1$$

I understood why we subtracted -1 because we are starting from base j=2 if i am right please mention that and explain me next doubt written below if not mention why we are using -1 at the end?

$$\sum^n_{j=2}(j-1)= n(n-1)/2$$

-

$$\sum_{j=1}^n j = 1+2+3+\ldots+(n-1)+n=\frac{n(n+1)}{2}$$ $$\sum_{j=2}^n j = 2+3+\ldots+(n-1)+n=2+3+\ldots+(n-1)+n+1-1=\frac{n(n+1)}{2}-1$$
You should take a look at the parenthesis. $\sum_{j=2}^{n}j-1 \neq \sum_{j=2}^{n}(j-1)$. – REr Jul 13 '14 at 12:44