a(n)+a(n+1)=2a(n+2),n=1,2,...
2a(n+2)-2a(n+1)=a(n)+a(n+1)-2a(n+1)=-[a(n+1)-a(n)],
b(n)=a(n+1)-a(n),n=1,2,...
b(n+1)=a(n+2)-a(n+1)=(1/2)[2a(n+2)-2a(n+1)]=(-1/2)[a(n+1)-a(n)]=(-1/2)b(n).
所以,{b(n)}是首项为b(1)=a(2)-a(1)=2-1=1,公比为(-1/2)的等比数列.
b(n)=(-1/2)^(n-1),n=1,2,...
(-1/2)^(n-1)=b(n)=a(n+1)-a(n).
a(n+1)-a(n)=(-1/2)^(n-1),
a(n)-a(n-1)=(-1/2)^(n-2),
...
a(3)-a(2)=(-1/2)^(2-1),
a(2)-a(1)=(-1/2)^(1-1).
上面n个等式,等号两边分别相加,有
a(n+1)-a(1)=(-1/2)^(n-1)+(-1/2)^(n-2)+...+(-1/2)+1=[(-1/2)^n-1]/[-1/2-1]=2[1-(-1/2)^n]/3,
a(n+1)=a(1)+2[1-(-1/2)^n]/3=1+2[1-(-1/2)^n]/3,n=1,2,...
a(n)=1+2[1-(-1/2)^(n-1)]/3,n=2,3,...
又,a(1)=1,
因此,有
a(n)=1+2[1-(-1/2)^(n-1)]/3,n=1,2,...