Mathematics OpenStudy (anonymous): OpenStudy (anonymous):

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

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? OpenStudy (roadjester):

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. OpenStudy (roadjester):

Samm, are you looking for 1 number or 5? OpenStudy (anonymous):

road: Neither, the question says to write it in expanded form. OpenStudy (roadjester):

Ok, my brain is pooped. OpenStudy (anonymous):

okay thanks

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