In the sequence {an}, it is known that A1 = 0, A2 = 6, and for any positive integer n, there is a (n + 2) = 5A (n + 1) - 6A (n) (1) Let BN = a (n + 1) - 2An, find the general term formula of BN (2) General term formula for an

In the sequence {an}, it is known that A1 = 0, A2 = 6, and for any positive integer n, there is a (n + 2) = 5A (n + 1) - 6A (n) (1) Let BN = a (n + 1) - 2An, find the general term formula of BN (2) General term formula for an

From a (n + 2) = 5A (n + 1) - 6A (n), we know that a (n + 2) - 2A (n + 1) = 3 [a (n + 1) - 2A (n)]
That is, B (n + 1) = 3bn, then {BN} is an equal ratio sequence. It is easy to get the general term formula of {BN} BN = 2 * 3 ^ n
From BN = a (n + 1) - 2An, we know that
a(n+1）-2an=2*3^n (*)
(*) is divided by 2 ^ (n + 1) at the same time
a(n+1）/2^(n+1)-an/2^n,=1.5^n (**)
Substituting CN = an / 2 ^ n into (* *) formula, C (n + 1) - CN = 1.5 ^ N accumulation is obtained
Cn-C1=1.5+1.5²+...+1.5^(n-1)
It is reduced to CN = 3 (1.5 ^ n-1) and replaced by CN = an / 2 ^ n
So an = 3 × 2 ^ n × (1.5 ^ n-1)

In the sequence {an}, A1 = 0, A2 = 2, a (n + 2) - 6A (n + 1) + 5An = 2 ^ n, find an
(n + 2), (n + 1) are subscripts

1. Let B [n] = a [n] + & _; × 2 ^ n, then B [1] = a [1] + & _; × 2 = & _; B [2] = a [2] + & _; × 2 ^ 2 = 10 / 3, and there is a [n] = B [n] - & _; × 2 ^ n, substituting a [n + 2] - 6A [n + 1] + 5A [n] = 2 ^ n and sorting, there is B [n + 2] - 6B [n + 1] + 5B [n]