Asked by kylie
Sum the series
1*2+2*3+3*4+ ... +2019*2020
1*2+2*3+3*4+ ... +2019*2020
Answers
Answered by
Reiny
General term = n(n+1) for n ∈ N
We want:
∑ n(n+1) for n=1 to 2019
= ∑ n^2 + n
= ∑ n^2 + ∑ n
= n(n+1)(2n+1)/6 + n(n+1)/2 , replace n with 2019 and evaluate
We want:
∑ n(n+1) for n=1 to 2019
= ∑ n^2 + n
= ∑ n^2 + ∑ n
= n(n+1)(2n+1)/6 + n(n+1)/2 , replace n with 2019 and evaluate
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