数列{an}满足a1=2,a2=5,a(n+2)=3a(n+1)-2an(1)求证:数列{a(n+1)-an}是等比数列 (2)求数列{an}的通项公式

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数列{an}满足a1=2,a2=5,a(n+2)=3a(n+1)-2an(1)求证:数列{a(n+1)-an}是等比数列(2)求数列{an}的通项公式数列{an}满足a1=2,a2=5,a(n+2)=

数列{an}满足a1=2,a2=5,a(n+2)=3a(n+1)-2an(1)求证:数列{a(n+1)-an}是等比数列 (2)求数列{an}的通项公式
数列{an}满足a1=2,a2=5,a(n+2)=3a(n+1)-2an
(1)求证:数列{a(n+1)-an}是等比数列 (2)求数列{an}的通项公式

数列{an}满足a1=2,a2=5,a(n+2)=3a(n+1)-2an(1)求证:数列{a(n+1)-an}是等比数列 (2)求数列{an}的通项公式
1.证明: a(n+2)-a(n+1)=2a(n+1)-2an a(n+1)-an =2an -2a(n-1) …… a3 - a2 =2a2 -2a1 全部相乘 a(n+2)-a(n+1) = 2^n(a2-a1)=3*2^n 数列{a(n+1)-an}=2^(n-1)*3 得证 2. 把上式相加 a(n+2)-a2=2a(n+1)-2a1 a(n+2)-2a(n+1)=1 由a(n+2)-a(n+1) = 2^n(a2-a1)=3*2^n a(n+1)=3*2^n-1 an=3*2^(n-1)-1

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