Question #ad76e

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Answer 1 · 2017-04-07T13:53:37

#:. sum_(k=1)^n (k^2-4k+3)=n/6(n-1)(2n-7).#

We will use # (1) : sum_(k=1)^n k^2=sumn^2=n/6(n+1)(2n+1),#

# (2) : sum n=n/2(n+1), and, sum a=an," for some const. a."#

Hence, #sum_(k=1)^n (k^2-4k+3),#

#=sum_(k=1)^nk^2-sum_(k=1)^n(4k)+sum_(k=1)^n3,#

#=sum n^2 - 4sum n + 3n,#

#=n/6(n+1)(2n+1)-4*n/2(n+1)+3n,#

#=n/6(n+1)(2n+1)-2n(n+1)+3n,#

#=n/6{(n+1)(2n+1)-12(n+1)+18},#

#=n/6{2n^2+3n+1-12n-12+18},#

#=n/6(2n^2-9n+7)#

#:. sum_(k=1)^n (k^2-4k+3)=n/6(n-1)(2n-7).#

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