Here's the question you clicked on:
Jonask
sigma
\[\sum_{k=1}^{49}(k+1)^3-k^3\] evaluate
no formular jus a series
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.
\[2^3-1^3+3^3-2^3+4^3-3^3+...+50^3-49^3\]
it appears to be a telescoping series
does it mean numbers are cancelling out
i think the anwers is \[-1^3+50^3\]
\[\sum_{k=1}^{49}(k+1)^3-k^3\] \[\sum_{k=1}^{49}(k+1)^3-\sum_{k=1}^{49}k^3\] \[\sum_{k=1+1}^{49+1}(k)^3-\sum_{k=1}^{49}k^3\] \[\sum_{k=2}^{50}k^3-\sum_{k=1}^{49}k^3=50^3-1^3\]