Why is $\sum_{i=0}^{n}1=(n+1)$?

Question

Why is $$\sum_{i=0}^{n}1=(n+1)?$$ I mean, as 1 does not depend on $i$, so shouldn't be the sum equals to 1 (as I was adding nothing, just keeping 1)?

You aren't "adding nothing", you are adding $1$ for each value of $i$ from $0$ to $n$. — Robert Israel, Jul 11 '14 at 04:02

score 9 · Accepted Answer · answered Jul 11 '14 at 00:20

9

In general, for any function $f$

$$\sum_{i=0}^n f(i) = f(0)+f(1)+f(2) + \dots + f(n)$$

Hence, if we let $f$ be the constant function $1$, then

$$\sum_{i=0}^n 1 = 1+ 1+ \dots + 1 = n+1$$

answered Jul 11 '14 at 00:20

Mastrel

score 3 · Answer 2 · answered Jul 11 '14 at 00:18

3

This summation means $1+1+1+1+\cdots+1$, where there are $n+1$ 1's in that string. This is because there are $n+1$ terms, and each one is 1.

answered Jul 11 '14 at 00:18

gregkow

2 Answers2