已知数列{an}满足a1=1,a2=2,a(n+2)=2/3a(n+1)+1/3an,求an
问题描述：
已知数列{an}满足a1=1,a2=2,a(n+2)=2/3a(n+1)+1/3an,求an
答
an=7/4 + 9/4 * (-1/3)^n
a(n+2)+a(n+1)/3=a(n+1)+an/3=a2+a1/3=7/3,
a(n+2)-a(n+1）=(-1/3)[a(n+1)-an]=[-1/3]^n*(a2-a1),
上下两个式子结合就可以得到an=7/4+9/4* (-1/3)^n懂了不∵3an+2 =2an+1 +an∴3an+2 -3an+1 = -(an+1-an)∴3(an+2 -an+1)= -(an+1-an)∴(an+2 -an+1)/ (an+1-an)=-1/3∴(an-an-1)/ (an-1-an-2)=-1/3(n≥3）∴{an - an-1}为等比例数列，公比为-1/3,首项为a2-a1=1