Summation Notation of i = 0 to N

Question

How do I compute the following $$ \sum_{i=0}^N 1 $$

If it were i instead of 1 then I would then have 0 + 1 + 2 ... + N. But, not sure how to do this. Thank you.

Answer 1

In general $\sum_{i=0}^Nf(i)=f(0)+\cdots+f(N)$, but in this case $f$ is the constant function $1$, so the sum is $N+1$. It's important to note the minimum $i$ is $0$ rather than $1$, to avoid an off-by-one error.

Answer 2

I claim $\sum_{i=0}^N 1=N+1,$ which I will prove through mathematical induction. When $N=0,$ $\sum_{i=0}^0 1=1=1+0.$ Now, assume the statement for $N=k.$ When $N=k+1,$ we have, by definition, $\sum_{i=0}^{k+1}1=(\sum_{i=0}^k 1)+1.$ Now, by our inductive assumption, we have $\sum_{i=0}^k 1=k+1,$ so the sum becomes $\sum_{i=0}^{k+1}1=(k+1)+1,$ thus proving the statement.