In the sequence an, A1 = 1, an + 1 = 2An + 2 ^ n, 1. Let BN = an / 2 ^ n-1. Prove that the sequence BN is an arithmetic sequence; can we not use the same division 2 ^ (n + 1) I mean to use the construction method to make a (n + 1) + k = 2 (an + k), but I use this method to find an = 3 * 2 ^ (n-1) - 2 ^ n, which is different from Baidu

In the sequence an, A1 = 1, an + 1 = 2An + 2 ^ n, 1. Let BN = an / 2 ^ n-1. Prove that the sequence BN is an arithmetic sequence; can we not use the same division 2 ^ (n + 1) I mean to use the construction method to make a (n + 1) + k = 2 (an + k), but I use this method to find an = 3 * 2 ^ (n-1) - 2 ^ n, which is different from Baidu

Divide both sides of the known equation by 2 ^ (n + 1) to get
a(n+1)/2^(n+1)=an/2^n+1/2 ,
So {an / 2 ^ n} is an arithmetic sequence whose first term is A1 / 2 = 1 / 2 and tolerance is 1 / 2,
So {an / 2 ^ n-1} is an arithmetic sequence with first term of - 1 / 2 and tolerance of 1 / 2
You can't make a (n + 1) + 2 ^ n = 2 (an + 2 ^ n), because the structures on both sides are not equal
That is to say, an + 2 ^ n on the right side should be equal to a (n + 1) + 2 ^ n (n + 1) on the left side