Question

In sigma n=11 then sigma ncube =

Answer 1

Answer:

Step-by-step explanation:

We know:

  Sigma (n^0) = n  

  Sigma (n^1) = n(n+1)/2

  Sigma (n^2) = n(n+1)(2n+1)/6

  Sigma (n^3) = (n(n+1)/2)^2

  ...

Is there an easy solution to find the result of Sigma (n^k), where k is  

a positive integer, like the right side of the equations above?

Thank you.