The known sequence an satisfies: an + 1-2an = 2 ^ n + 1, and A1 = 2 (1). It is proved that {an / 2 ^ n} is the key to find the sequence an of arithmetic sequence (2)

I don't know if your 2 ^ n + 1 is 2 ^ (n + 1)
(1) Divide both sides of an + 1-2an = 2 ^ n + 1 by 2 ^ (n + 1) to get a (n + 1) / 2 ^ (n + 1) - an / 2 ^ n = 1
Because A1 / 2 = 1, the sequence {an / 2 ^ n} is an arithmetic sequence with 1 as the first term and 1 as the tolerance
Then there is an / 2 ^ n = 1 + (n-1) * 1 = n
So an = n * 2 ^ n
(2) From (1), we know that Sn = 1 * 2 + 2 * 2 ^ 2 + 3 * 2 ^ 3 +.. + n * 2 ^ n
2sn=1*2^2+2*2^3+3*2^4+..+（n-1）2^n+n*2^（n+1）
So sn-2sn = 1 * 2 + (2 ^ 2 + 2 ^ 3 + 2 ^ 4.. + 2 ^ n) - n * 2 ^ (n + 1)
That is Sn = - (2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 +.. + 2 ^ n) + n * 2 ^ (n + 1)
sn= -[1*（1-2^n）]+n*2^（n+1）
=-1-2^n+n*2^（n+1）
=（4n-1）*2^（n-1）- 1

In the positive term sequence an, A1 = 1, an + 1 - √ an + 1 = an + √ an. Prove that the sequence an is an arithmetic sequence and find the general term an
It is proved that √ an is an arithmetic sequence

an+1-√an+1=an+√an
We get an + 1-an = √ an + 1 + √ an
That is (√ an + 1 + √ an) (√ an + 1 - √ an) = √ an + 1 + √ an
Then √ an + 1 - √ an = 1
So {an} is an arithmetic sequence with the first term of √ A1 = 1 and the tolerance of 1
Then √ an = n
So an = n ^ 2

It is known that the sequence an satisfies A1 = 3, an + 1 = 2An + 2 ^ n (1) to prove that the sequence [an / 2 ^ n] is an arithmetic sequence (2) to find the general formula of an

(1) Proof: a (n + 1) = 2An + 2 & # 8319; both sides of the equation are divided by 2 ^ (n + 1) a (n + 1) / 2 ^ (n + 1) = an / 2 & # 8319; + 1 / 2A (n + 1) / 2 ^ (n + 1) - an / 2 & # 8319; = 1 / 2, which is the fixed value. A1 / 2 = 3 / 2, and the sequence {an / 2 & # 8319;} is an arithmetic sequence with 3 / 2 as the first term and 1 / 2 as the tolerance. (2) an / 2 & # 8319; = (3 / 2) + (n -

Given that the sequence {an} is an arithmetic sequence, A1 = 1, tolerance is 2, and that the sequence {BN} is an equal ratio sequence, and B1 = A1, B2 (a2-a1) = B1, find the general term formula 2 of the sequence {an}, {BN}. "Let CN = an / BN, find the first n term and Sn of {CN}

A1 = 1, A2 = 3, so an = 2n-1b1 = 1, B2 = 0.5, so an = (0.5) ^ (n-1) = 2 ^ (1-N) 2. CN = an / BN = (2n-1) * 2 ^ (n-1) Sn = 1 * 2 ^ 0 + 3 * 2 ^ 1 + 5 * 2 ^ 2 + +(2 (n-1) - 1) * 2 ^ ((n-1) - 1) + (2n-1) * 2 ^ (n-1). Formula 12sn = 1 * 2 ^ 1 + 3 * 2 ^ 2 + 5 * 2 ^ 3 + +(2(n-1)-1)*2^(n-1)...

The known sequence an satisfies: an + 1-2an = 2 ^ n + 1, and A1 = 2 (1). It is proved that {an / 2 ^ n} is the key to find the sequence an of arithmetic sequence (2)