## anonymous 3 years ago sigma

1. anonymous

What?

2. anonymous

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

3. anonymous

no formular jus a series

4. 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.

5. anonymous

$2^3-1^3+3^3-2^3+4^3-3^3+...+50^3-49^3$

6. TuringTest

it appears to be a telescoping series

7. anonymous

does it mean numbers are cancelling out

8. anonymous

i think the anwers is $-1^3+50^3$

9. amistre64

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