a(n- 1)-a(n-2)=3 2^(2n-5)

...

a(2)-a( 1)=3 2^ 1

a( 1)=2

All kinds of accumulation, there are.

When n≥2, a (n) = 3 [21+23+...+2 (2n-3)]+2.

=3 2[ 1-4^n]/( 1-4)+2

=2 4^n

When n= 1, a(n)=2.

To sum up, a (n) = 2.4 n

b(n)=na(n)=2n 4^n

So the sum of the first n terms of b(n) is

s(n)= 2 ^ 4+4 ^ 4？ +6 4? +...+2n 4^n...①

4S(n)= 2 ^ 4？ +4 4? +...+2(n- 1)4^n+2n 4^(n+ 1)...②

①-②, yes

-3S(n)= 2 ^ 4+2(4？ +4? +...+4^n)-2n 4^(n+ 1)

=8+2 [4? ( 1-4^n- 1)/ 1-4]-2n 4^(n+ 1)

= 8+32/3[4^(n- 1)- 1]-2n 4^(n+ 1]

=8/3 4^n-8/3-8n 4^n

So s (n) = (8n/3-8/9) 4 n-8/9.

Hope to adopt