How do I solve this Sigma Problem?

Question

Given that:
#sum_(k=1)^n k=(n(n+1))/2# and # sum_(k=1)^n k^2=(n(n+1)(2n+1))/6 #

Evaluate
#sum_(k=1)^20 (2k^2+5k-1).#

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Answer 1 · 2018-03-11T06:26:31

6770

Sum = #sum_(k=1)^20 (2k^2+5k-1)#

Apply linearity.

Sum = #2sum_(k=1)^20 k^2 + 5sum_(k=1)^20 k - sum_(k=1)^20 1#

We are told that:

#sum_(k=1)^n k^2 = (n(n+1)(2n+1))/6#

and

#sum_(k=1)^n k = (n(n+1))/2#

Replacing the above into "Sum" with #n=20#

Sum#= 2xx(20(20+1)(2*20+1))/6 +5xx (20(20+1))/2 -20xx1#

#= 2xx(20*21*41)/6 + 5xx (20*21)/2 -20#

#= 5740 +1050 -20#

#=6770#