# What does $\sum_{i=1}^{10} 2$ mean exactly?

Suppose I have $$\large\sum_{i=1}^{10} 2.$$ Do I just add $2$ to itself $10$ times? I have worked on more complex ones with $n$ and such in the place where the $2$ is, but I have never done it when there is just a number there.

• You've got good intuition! Many people would just assume it's equal to 2. Apr 28, 2015 at 19:16

You are exactly right. Many summations are of the following form: $$\sum_{i=1}^n c\cdot f(i) = c\cdot \sum_{i=1}^n f(i)$$ Where $f(i)$ is some function of $i$. Notice in this case, $f(i) = 1$, and you are left with $$\sum_{i=1}^{10} 2\cdot f(i) = 2\cdot\sum_{i=1}^{10} 1 = 2\cdot 10$$

• Thank you for the further explanation! Apr 28, 2015 at 19:04
• Nice explanation. +1 Apr 28, 2015 at 19:08
• But where does he explain how $\sum_{i=1}^{10} 1 = 10$? Apr 30, 2015 at 21:16
• @StevenGregory, I tailored the post to the OP. He demonstrated his understanding of the notation, and I omitted this from the explanation. Apr 30, 2015 at 21:18

Read it as $\large\sum_{i=1}^{10} a_i$ where $a_i = 2$ for all $i$.

So it's

$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} =$

$2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2$

The summation of a constant is just the constant time the bound. so in your case the result would be 20. $$\sum_{i=1}^{10} 2 = 2\cdot 10 = 20$$ http://www.psychstat.missouristate.edu/introbook/sbk12m.htm

• Alright thank you, I was assuming that but I was not exactly positive and my teacher wouldn't be able to respond to my email until tomorrow. Apr 28, 2015 at 19:03

This is another example of how notation can needlessly complicate things. It would be better to write $2 \times 10$ or $10 \times 2$ (too early in the morning for me to argue about commutativity).

If you have any lingering doubts, try this in Wolfram Alpha: Sum[2, {n, 1, 10}].

But don't think that a constant summand in an iterated sum is always pointless. Things like the prime counting function can be defined with a constant summand of 1: $$\pi(n) = \sum_{p \leq n, p \textrm{ prime}} 1.$$