# Partial sum of 1-x

• Apr 2nd 2011, 07:44 AM
iva
Partial sum of 1-x
On WolframAlpha I see that the sum of x+1 is given by

$\displaystyle\sum_{i=0}^{n} -\frac{1}{2}(n-2)(n+1)$

But if I want to start at i=1 this sum is no longer valid!! Was a surprise at first but i see now it makes sense, but then what is the way to adjust these partial sum formulas for starting at different numbers, or do I have to think of a totally new formula when i want to do something like this?

Thank you
• Apr 2nd 2011, 08:05 AM
TheEmptySet
Quote:

Originally Posted by iva
On WolframAlpha I see that the sum of x+1 is given by

$\displaystyle\sum_{i=0}^{n} -\frac{1}{2}(n-2)(n+1)$

But if I want to start at i=1 this sum is no longer valid!! Was a surprise at first but i see now it makes sense, but then what is the way to adjust these partial sum formulas for starting at different numbers, or do I have to think of a totally new formula when i want to do something like this?

Thank you

First something is weird with your indexes. You are summing over i but there are know i's in your formula!

I assume you mean something like this

Expand the sum out to get

$\displaystyle -\frac{1}{2}\left( \sum_{k=0}^{n}k^2- \sum_{k=0}^{n}k-\sum_{k=0}^{n}2\right)$

The sum or squares and the sum or the first n integers are known formula's