Help with finding the value of n

[angelcause]

Hello, I am new with sigma and can't figure the below out, please help me.

\(\displaystyle \sum_{i=0}^{n}\)i=n(n+1)/2Thanks a million.

 

[Evgeny.Makarov]

Welcome to the forum!

This sequence is a special case of an arithmetic progression.

\(\displaystyle \sum_{i=1}^n i=1+2\dots+(n-1)+n=(1+n)+(2+n-1)+\dots=(n+1)+(n+1)+\dots\)

If $n$ is even, then the last sum has $n/2$ terms, so the result is $(n+1)n/2$. If $n$ is odd, then there are $(n-1)/2$ terms equal to $n+1$ and there remains the middle term of the original sequence, which is $(n+1)/2$. The total sum is $(n+1)(n-1)/2+(n+1)/2=(n+1)n/2$.

The link above has a simpler proof.

 

[angelcause]

Thank you but the value for i is 0, can u please get it step by step

 

[Evgeny.Makarov]

the value for i is 0

In this sum, $i$ does not have a fixed value. Instead, the value of $i$ ranges from $0$ to $n$. But note that
\[
\sum_{i=0}^ni=0+1+\dots+(n-1)+n=1+\dots+(n-1)+n=\sum_{i=1}^ni.
\]

 

[soroban]

Hello, I am new with sigma and can't figure the below out, please help me.

[tex]\displaystyle \sum_{i=0}^{n}i \;=\; \frac{n(n+1)}{2}[/tex]

Thanks a million.

It says, "the sum of the [tex]i's[/tex] as [tex]i[/tex] goes from 0 to [tex]n[/tex]".

[tex]\displaystyle \sum^n_{i-0}i \;=\;0 + 1 + 2 + 3 + 4 + \cdots + n[/tex]