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

(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