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设数列an满足 a1=5/6,且an=1/3a(n-1)+1/3 求证 数列an-1/2为等比数列

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设数列an满足 a1=5/6,且an=1/3a(n-1)+1/3 求证 数列an-1/2为等比数列
并求an的通项公式与前n项和Sn
设数列an满足 a1=5/6,且an=1/3a(n-1)+1/3 求证 数列an-1/2为等比数列
an=(1/3)a(n-1)+1/3
an -1/2 = (1/3)[a(n-1) -1/2]
(an -1/2)/[a(n-1) -1/2] =1/3
=>{an-1/2}为等比数列
(an -1/2)/[a(n-1) -1/2] =1/3
(an -1/2)/[a1 -1/2] =(1/3)^(n-1)
an-1/2 = (1/3)^n
an = 1/2 + (1/3)^n
Sn =a1+a2+...+an
= n/2 + (1/2)(1-(1/3)^n)