5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ You've got good intuition! Many people would just assume it's equal to 2. $\endgroup$ Apr 28, 2015 at 19:16

4 Answers 4

5
$\begingroup$

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 $$\sum_{i=1}^{10} 2\cdot f(i) = 2\cdot\sum_{i=1}^{10} 1 = 2\cdot 10$$

$\endgroup$
4
  • $\begingroup$ Thank you for the further explanation! $\endgroup$ Apr 28, 2015 at 19:04
  • $\begingroup$ Nice explanation. +1 $\endgroup$ Apr 28, 2015 at 19:08
  • $\begingroup$ But where does he explain how $\sum_{i=1}^{10} 1 = 10$? $\endgroup$ Apr 30, 2015 at 21:16
  • $\begingroup$ @StevenGregory, I tailored the post to the OP. He demonstrated his understanding of the notation, and I omitted this from the explanation. $\endgroup$ Apr 30, 2015 at 21:18
4
$\begingroup$

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

So it's

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Apr 28, 2015 at 19:03
1
$\begingroup$

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: $$\pi(n) = \sum_{p \leq n, p \textrm{ prime}} 1.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .