Mathematics
OpenStudy (anonymous):

OpenStudy (anonymous):

$\sum_{m=1}^{5}(2m ^{2}+1)$

Uh, expanded form? As in the sequence? You only have 5 terms so it's a sequence but you have sigma which leads me to believe that it's a sum

OpenStudy (anonymous):

You're looking for the sum of (2m^2+1) for each of m=1 through m=5. So it will start with $$(2\cdot1^2+1)$$, can you take it from there?

3,3+9,12+19,etc

OpenStudy (anonymous):

I'm just going to write it out to make sure this is clear. You're looking for:$(2\cdot1^2+1)+(2\cdot2^2+1)+(2\cdot3^2+1)+(2\cdot4^2+1)+(2\cdot5^2+1)$

OpenStudy (anonymous):

Do you understand how I got that?

OpenStudy (anonymous):

nbouscal is this he answer 3,9,19,33,51 or do i add them equalling 115

OpenStudy (anonymous):

They are summed, but the question is not asking for the solution, it is asking for the expanded form of it, which is what I wrote in my previous reply.

Samm, are you looking for 1 number or 5?

OpenStudy (anonymous):

road: Neither, the question says to write it in expanded form.