Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):

rewrite in expanded form please

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

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
2828838399292: The map above traces the journey of
3 hours ago 0 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!