Question

设数列{an}的前n项和为sn,且sn=n^2-4n+4,求数列｛an｝的通项公式

设数列{an}的前n项和为sn,且sn=n^2-4n+4,1.求数列｛an｝的通项公式2.设bn=an/2n,数列｛bn｝的前n项和为TN

Answer 1

an=sn-sn-1=2n-5
bn=1-5/2n Tn=b1+b2+b3+.......+bn=n-5/2(1+2+3+....+n)=n-5/2(n ^2+1/2) =-(5n^ 2+n/4)

Answer 2

n=1时，a1=s1=1
n>=2时 an=sn-s(n-1）=2n-5