Question

What is the partial sum of ?

Answer 1

`n(n+1)-20`

Explanation:

`sum_k^n2k=2sum_k^n k`

but

`1+2+3+4+5+6+cdots+n =(n(n+1))/2`

then

`sum_{k=m}^n k = sum_{k=1}^nk-sum_{k=1}^{m-1}k =(n(n+1))/2-(m(m-1))/2`

finally

`sum_{k=5}^n 2k = 2((n(n+1))/2-(5cdot 4)/2) = n(n+1)-20`