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Mathematics 15 Online
OpenStudy (anonymous):

sigma

OpenStudy (anonymous):

What?

OpenStudy (anonymous):

\[\sum_{k=1}^{49}(k+1)^3-k^3\] evaluate

OpenStudy (anonymous):

no formular jus a series

OpenStudy (anonymous):

That series is just saying when you plug in k = 1, 2, 3, 4....49 and add them up, there should be a formula for the series to make the calculation easier. It is possible to calculate every term.

OpenStudy (anonymous):

\[2^3-1^3+3^3-2^3+4^3-3^3+...+50^3-49^3\]

OpenStudy (turingtest):

it appears to be a telescoping series

OpenStudy (anonymous):

does it mean numbers are cancelling out

OpenStudy (anonymous):

i think the anwers is \[-1^3+50^3\]

OpenStudy (amistre64):

\[\sum_{k=1}^{49}(k+1)^3-k^3\] \[\sum_{k=1}^{49}(k+1)^3-\sum_{k=1}^{49}k^3\] \[\sum_{k=1+1}^{49+1}(k)^3-\sum_{k=1}^{49}k^3\] \[\sum_{k=2}^{50}k^3-\sum_{k=1}^{49}k^3=50^3-1^3\]

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