MHB Help with finding the value of n

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The discussion revolves around understanding the summation formula for the series from 0 to n, specifically $$\sum_{i=0}^{n} i = \frac{n(n+1)}{2}$$. Participants clarify that the sum includes all integers from 0 to n, and they explain the derivation of the formula using properties of arithmetic progressions. The conversation highlights that the value of i is not fixed but varies within the defined range. A step-by-step explanation is requested to further clarify the summation process. The overall focus is on providing a clear understanding of the sigma notation and its application in calculating the sum.
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Hello, I am new with sigma and can't figure the below out, please help me.

$$\sum_{i=0}^{n}$$i=n(n+1)/2Thanks a million.
 
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Welcome to the forum!

This sequence is a special case of an arithmetic progression.

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

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.
 
Thank you but the value for i is 0, can u please get it step by step
 
angelcause said:
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
\[
\sum_{i=0}^ni=0+1+\dots+(n-1)+n=1+\dots+(n-1)+n=\sum_{i=1}^ni.
\]
 
angelcause said:
Hello, I am new with sigma and can't figure the below out, please help me.

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

Thanks a million.
It says, "the sum of the i's as i goes from 0 to n".

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