It is known that the sum of the first n terms of the sequence {an} is Sn, and a1 + 2A2 + 3a3

It is known that the sum of the first n terms of the sequence {an} is Sn, and a1 + 2A2 + 3a3

It is known that the sum of the first n terms of the sequence {an} is Sn, and a1 + 2A2 + 3a3 + +nan=（n-1）Sn+2n
Find an
a1+2a2+3a3+… +nan=（n-1）sn+2n（n€N*）
(1) First, A1
n=1,
∴ a1=2
(2) Using the recursive formula:
∵ a1+2a2+3a3+…… +(n-1)a(n-1)+nan=(n-1)Sn+2n ①
∴ a1+2a2+3a3+…… +(n-1)a(n-1) =(n-2)S(n-1)+2(n-1) ②
①-②：
nan =(n-1)Sn-(n-2)S(n-1)+2n
That is n [S (n) - S (n-1)] = (n-1) Sn - (n-2) s (n-1) + 2
∴ S(n)=2S(n-1)+2
∴ S(n)+2=2[S(n-1)+2]
{Sn + 2} is an equal ratio sequence with S1 + 2 = 2 + 2 = 4 as the first term and 2 as the common ratio,
∴ Sn+2=4*2^n=2^(n+1)
∴ Sn=-2+2^(n+1)
When n ≥ 2, an = SN-S (n-1) = 2 ^ (n + 1) - 2 ^ (n) = 2 ^ n
N = 1 also satisfies the above equation
∴ an=2^n

The known sequence {an} satisfies a1 + 2A2 + 3a3 + +Nan = n (n + 1) (n + 2)=______ ．

∵a1+2a2+3a3+… +nan=n（n+1）（n+2），①∴a1+2a2+3a3+… +(n-1) an-1 = (n-1) n (n + 1), ② (1) - ②, we get Nan = 3N (n + 1), х an = 3N + 3. х Sn = a1 + A2 + a3 + +an=（3×1+3）+（3×2+3）+（3×3+3）+… +（3n+3）=3（1+2+3+… +n) So the answer is: 3N2 + 9n2

Let the sum of the first n terms of the sequence an be Sn, a1 + 2A2 + 3a3 + +nan=(n-1)Sn+2n
Take out A1, A4, A7. A (3n-2) in {an}, and the rest of the order remains unchanged to form {BN}. If the sum of the first n terms of {BN} is TN, it is proved that 12 / 5 ＜ t (n + 1) / TN ≤ 11 / 3

From a1 + 2A2 + 3a3 + +Nan = (n-1) Sn + 2n: when n = 1: A1 = (1-1) S1 + 2, the solution is: A1 = 2; when n = 2: a1 + 2A2 = (2-1) S2 + 4, that is, 2 + 2A2 = (2 + A2) + 4, the solution is: A2 = 4 +nan=(n-1)Sn+2na1+2a2+3a3+…… +nan+(n+1)a(n+1)=[(n+1...

The sequence {an} satisfies an = 3an-1 + 3 ^ n-1 (n ≥ 2), where A4 = 365. Prove that the sequence {an-3 ^ n} is not an arithmetic sequence

An=3An-1+3^(n-1)
Divide both sides by 3 ^ (n-1)
3*An/3^n=3An-1/(3^n-1)+1
An/3^n=An-1/(3^n-1)+1/3
Then an / 3 ^ n is an arithmetic sequence with a tolerance of 1 / 3
Then an / 3 ^ n = A1 / 3 + (n-1) / 3
Then an = [A1 / 3 + (n-1) / 3] * 3 ^ n
Then A4 = [A1 / 3 + (4-1) / 3] * 3 ^ 4 = 365, the solution is A1 = 284 / 27
Then an = [284 / 81 + (n-1) / 3] * 3 ^ n = (n + 257 / 27) * 3 ^ (n-1)
Then an-3 ^ n = (n + 257 / 27) * 3 ^ (n-1) - 3 ^ n = (n + 176 / 27) * 3 ^ (n-1)
Then an-3 ^ n - [a (n-1) - 3 ^ (n-1)]
=(n+176/27)*3^(n-1)-(n-1+176/27)*3^(n-2)
=(2n + 379 / 27) * 3 ^ (n-2) is not a constant, but a function of n,
So an-3 ^ n is not an arithmetic sequence