0
$\begingroup$

Disclaimer: This is for homework, but I'm just stuck on this one small part of a larger problem.

I'm having trouble figuring out how to get the following summation in closed form.

$$\sum_{j=1}^i 4ij$$

Since the index is j, am I able to move the i out of the summation as a constant the way I am with the 4? If I do that, do I need to square the i because it is the stopping point of the summation?

Am I able to separate the summation into two, like this?

$$\left(\sum_{j=1}^i 4i\right)\left(\sum_{j=1}^i j\right)$$

$\endgroup$
3
$\begingroup$

The $i$ is indeed a constant, so you treat it exactly as like the $4$:

$$\sum_{j=1}^i4ij=4i\sum_{j=1}^ij=4i\left(\frac{i(i+1)}2\right)=2i^2(i+1)\;.$$

It might help you to write out some terms: the sum is

$$4i\cdot1+4i\cdot2+4i\cdot3+\ldots+4i\cdot i\;,$$

which clearly has a factor of $4i$ in each term that can be factored out to yield

$$4i(1+2+3+\ldots+i)\;.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.