数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,n属于N*.求(1)令bn=a(n+1)-an，证明{bn}是等比数列。(2)求{an}通项公式。求(1)令bn=a(n+1)-an，证明{bn}是等比数列。(2)求{an}通项公式。

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