设A1=2,A2=4,数列{Bn}满足:Bn=A(n+1) –An,B(n+1)=2Bn+2.

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设A1=2,A2=4,数列{Bn}满足:Bn=A(n+1) –An,B(n+1)=2Bn+2.
(1) 求证:数列{ Bn+2}是等比数列(要指出首项与公比)
(2) 求数列{ An}的通项公式.

(1) B(n+1)=2B(n)+2
=>B(n+1)+2 = 2( B(n)+2 )
所以:B(n)+2 是等比数列
公差为2,首项 B1+2 = 4
(2) B(n) = A(n+1) - A(n)
B(n-1) = A(n) - A(n-1)
.
B(1) = A(2) - A(1)
上面n个式子相加可得
B(1)+B(2)+...+B(n) = A(n+1)-A(1)
=>( B(1)+2 )+( B(2)+2 )+ ...+( B(n)+2 )
= A(n+1) - A(1) + 2*n
=>4 + 8 + 16 + ...+ 4*2^(n-1)
= A(n+1) - 2 + 2*n
=> A(n+1) = 2^(n+2) - 2n - 2
=> A(n) = 2^(n+1) - 2n

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