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
Øko
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#

Cesareo R.
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#

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