How to evaluate this sum?

Evaluate 99k=1k(k21)

Given that:
nk=1k=n(n+1)2

and

nk=1k3=[nk=1k]2

2 Answers
Mar 12, 2018

S99=24497550

Explanation:

We want to evaluate

S99=99k=1k(k21)

Let's take the general example

Sn=nk=1k(k21)=nk=1k3k=nk=1k3nk=1k

Using

  • nk=1k=n(n+1)2
  • nk=1k3=(nk=1k)2=(n(n+1)2)2

Thus

Sn=(n(n+1)2)2n(n+1)2

For n=99

S99=(99(99+1)2)299(99+1)2

S99=(99002)299002

S99=99002198004=24497550

Mar 12, 2018

24497550

Explanation:

Given that:

nk=1k=n(n+1)2

and

nk=1k3=[nk=1k]2

follows

nk=1k3nk=1k=nk=1(k3k)=nk=1k(k21)

hence

nk=1k(k21)=(n(n+1)2)2n(n+1)2=

=14(n2+n2)n(n+1)

and for n=99 gives

nk=1k(k21)=24497550