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
Evaluate
Given that:
and
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
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
Explanation:
Given that:
and
follows
hence
and for