How to evaluate this sum?

Evaluate sum_(k=1)^99 k(k^2-1)

Given that:
sum_(k=1)^n k = (n(n+1))/2

and

sum_(k=1)^n k^3 = [sum_(k=1)^n k ]^2

2 Answers
Mar 12, 2018

S_99=24497550

Explanation:

We want to evaluate

S_99=sum_(k=1)^99k(k^2-1)

Let's take the general example

S_n=sum_(k=1)^nk(k^2-1)=sum_(k=1)^nk^3-k=sum_(k=1)^nk^3-sum_(k=1)^nk

Using

  • sum_(k=1)^nk=(n(n+1))/2
  • sum_(k=1)^nk^3=(sum_(k=1)^nk)^2=((n(n+1))/2)^2

Thus

S_n=((n(n+1))/2)^2-(n(n+1))/2

For n=99

S_99=((99(99+1))/2)^2-(99(99+1))/2

S_99=((9900)/2)^2-(9900)/2

S_99=(9900^2-19800)/4=24497550

Mar 12, 2018

24497550

Explanation:

Given that:

sum_(k=1)^n k = (n(n+1))/2

and

sum_(k=1)^n k^3 = [sum_(k=1)^n k ]^2

follows

sum_(k=1)^n k^3 - sum_(k=1)^n k = sum_(k=1)^n(k^3-k) = sum_(k=1)^nk(k^2-1)

hence

sum_(k=1)^nk(k^2-1) = ((n(n+1))/2)^2-(n(n+1))/2=

=1/4(n^2+n-2)n(n+1)

and for n=99 gives

sum_(k=1)^nk(k^2-1) =24497550