Question

已知数列{an}的前n项和为Sn,通项an满足Sn+an=1/2(n²+3n-2)

求通项公式an

Answer 1

解答为：Sn+an=1/2*(n^2+3n-2)...........................(1)
S(n-1)+a(n-1)=1/2*[(n-1)^2+3(n-1)-2]............(2)
(1)-(2):
an+an-a(n-1)=n+1
2an-a(n-1)=n+1
2an-n-1=a(n-1)
即:2(an-n)=a(n-1)-(n-1)
即:(an-n)/[a(n-1)-(n-1)]=1/2
∴{an-n}是公比为1/2的等比数列
令{an-n}={bn}
由题意:
2a1=1/2*(1+3-2)
∴a1=1/2
∴b1=a1-1=-1/2
∴bn=(-1/2)*(1/2)^(n-1)
=-(1/2)^n=an-n
∴an=-(1/2)^n+n,n∈N+