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Mathematics 7 Online
OpenStudy (anonymous):

find the sum of the sequence (look below)

OpenStudy (anonymous):

\[\sum_{k=1}^{4} \left( k ^{3} +4\right)\]

OpenStudy (anonymous):

write it out

OpenStudy (primeralph):

put 1 for k, then 2, then 3 , then 4. Add all the answers together

OpenStudy (anonymous):

wow it was that simple? looked more complex so the answer i got was 5, 12, 31, 68

OpenStudy (primeralph):

how did you get 31?

OpenStudy (primeralph):

and 68?

OpenStudy (anonymous):

I just plug in 3 and 4 for k

OpenStudy (anonymous):

and all I have to do is add them up right?

OpenStudy (primeralph):

yeah, plug in 1,2,3,4. Add up the results. But, 31, 68 ain't part of the 'plug in results'

OpenStudy (anonymous):

really? i got 31 from plugging in 3 and 68 by plugging in 4?

OpenStudy (primeralph):

3^2 +4 = 9+4 = 13 or is it ^3 you wrote?

OpenStudy (primeralph):

if it's k^3, then you're right..........couldn't see the power clearly....

OpenStudy (anonymous):

yeah its ^3 so do I just add them all up for the sum

OpenStudy (primeralph):

yeah. That's all.

OpenStudy (anonymous):

thanks again!! You really are a big help!

OpenStudy (primeralph):

yeah, you're very welcome.

OpenStudy (primeralph):

you could also.post a testimonial for me.........(-_-)

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