In the sequence {an}, A1 = 1, A2 = 6, a_ n+2=A_ N + 1-an, then A2010 is equal to?

In the sequence {an}, A1 = 1, A2 = 6, a_ n+2=A_ N + 1-an, then A2010 is equal to?

A1 = 1 ,A 2 = 6 ,A3 = 5 ,A4 = -1 ,A5 = -6 ,A6 = -5
A_ n+6 = A_n
6-bit one cycle
Because 2010 △ 6 = 335, there is no remainder
So A2010 = A6 = - 5

Let an satisfy A1 = 2 an + 1-an = 3-2 ^ 2N-1
1. Find the general term of the sequence
2. Let BN = n * an find the sum of the first n terms of BN

(1) According to the meaning of the title, an = (an-an-1) + (an-1 - An-2) + +（A2 - A1）+A1
=3-2^(2n-3) + 3-2^(2n-5) +… +(3-2^3)+ 2
Then, the group summation method is used
=3n - 【2^(2n-3)+2^(2n-5)+… 2^3+2】
=3n-2*(1-4^n)\(1-4)
=2*(1-4^n)\3 + 3n
That is, an = - 2 * (4 ^ n-1) / 3 + 3N
(2)Bn=n*An==-2n*(4^n-1)\3 + 3n^2
The same as (1): using group summation
Help me add points, I'm a rookie

Let {an} satisfy A1 = 2, an + 1-an = 3.2 ^ (2n-1)
Let BN = n · an, find the first n terms and Sn of sequence {BN}