How do you simplify #\sum _ { k = 1} ^ { n } ( 3k ^ { 2} + 2k )#?

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Answer 1 · 2017-11-08T13:00:51

The answer is #=n/2(n+1)(2n+3)#

Let's rewrite the summation

#sum_(k=1)^n3k^2+sum_(k=1)^n2k#

#=3sum_(k=1)^nk^2+2sum_(k=1)^nk#

#=3*n/6(n+1)(2n+1)+2*n/2(n+1)#

#=n/2(n+1)(2n+1)+n(n+1)#

#=n(n+1)(1/2(2n+1)+1)#

#=n(n+1)(n+1/2+1)#

#=n/2(n+1)(2n+3)#