数列an满足A1=1,A(n+2)=2A(n+1)-A(n+2)求1、设Bn=A(n+1)-An求证Bn是等差数列.2、求An的通项公式.
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数列an满足A1=1,A(n+2)=2A(n+1)-A(n+2)求1、设Bn=A(n+1)-An求证Bn是等差数列.2、求An的通项公式.数列an满足A1=1,A(n+2)=2A(n+1)-A(n+2
数列an满足A1=1,A(n+2)=2A(n+1)-A(n+2)求1、设Bn=A(n+1)-An求证Bn是等差数列.2、求An的通项公式.
数列an满足A1=1,A(n+2)=2A(n+1)-A(n+2)求1、设Bn=A(n+1)-An求证Bn是等差数列.2、求An的通项公式.
数列an满足A1=1,A(n+2)=2A(n+1)-A(n+2)求1、设Bn=A(n+1)-An求证Bn是等差数列.2、求An的通项公式.
1)略
2)BN=n^2-7n-8
【解析】bn=A(n+1)-An,b(n+1)=A(n+2)-A(n+1),则
A(n+2)-2A(n+1)+An=2n-6=b(n+1)-bn,于是b(n)-b(n-1)=2n-8
b(n-1)-b(n-2)=2n-10
……
b4-b3=0
b3-b2=-2
b2-b1=-4
b1=A2-A1=-14
两边相加得【累加法】
b(n)=-14+[-4-2+0+2+4+……+2(n-4)]=-14-6+2*(n-4)(n-3)/2=n^2-7n-8
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