Question

已知An前n项和为Sn,Sn=n^2+1/2An 求证A(n+1)+An=4n+2 求An

Answer 1

1.
Sn=n^2+An/2
S(n+1)=(n+1)^2+A(n+1)/2
S(n+1)-Sn=(n+1)^2+A(n+1)/2-n^2-An/2=A(n+1
n^2+2n+1-n^2=An/2+A(n+1)/2
A(n+1)+An=4n+2
2.
A(n+1)+An=4n+2
A(n+2)+A(n+1)=4n+6
两式想减A(n+2)-An=4
所以奇数项成等差数列,偶数项也成等差数列
S1=1^2+1/2A1=A1
A1=2
A2+A1=4+2
A2=6-2=4
可以发现奇数项和偶数项合并后还是等差数列,但是公差变为2
An=2+2（n-1）=2n