(1) a [n]-1= 2 (a [n-1]-1)+2n (the left and right sides of the equation are divided by 2n at the same time) = > (a [n]-1)/(.

That is: b [n] = b [n-1]+1= > b [n]-b [n-1] =1(constant), so {b[n]} is arithmetic progression with a tolerance of.

(2) b [1] = (a [1]-1)/2 = 2, then b [n] = (n-1) d+b [1] = n+. Namely: (a [n]- 1)/(2 n) = n+ 1,

So a [n] = (n+ 1) (2 n)+ 1.

So s [n] = 2 (21)+3 (22)+...+(n+1) (2 n)+n one.

Then 2s [n] = 2 (2 2)+...+n (2n)+(n+1) (2n+1)+2n.

Two MINUS one:

s[n]=-2(2^ 1)-(2^2)+……(2^n)+(n+ 1)(2^n+ 1)+n

=-4+4 [1-(2n-1)]/(1-2)+(n+1) (2n+1)+n (middle n-65438

=-4-4+(2^n+ 1)+(n+ 1)(2^n+ 1)+n

=-8+(n+2)(2^n+ 1)+n

So s [n] = (n-8)+(n+2) (2 n+ 1)