an+ 1-an=3*2^(2n- 1)

an-an- 1=3*2^(2n-3)

an- 1-an-2=3*2^(2n-5)

......

a2-a 1=3*2

Add left and right to get a (n+1)-a1= 3 * (21+2 3+2 5+2 7+...+2 (2n-1)).

21+23+25+27+...+2 (2n-1) is the sum of geometric series, and the first term of the common ratio is 2.

Let TN = 21+23+25+27+...+2 (2n-1).

=2×( 1-4^n)/( 1-4)

=2*(4^n- 1)/3

Then you can get the above formula:

a(n+ 1)-a 1=3*(2*(4^n- 1)/3)

=2*(4^n- 1)

a 1=2

Get a (n+1) = 2× 4 n.

=2^(2n+ 1)

That is, an = 2 (2n- 1)

2.

bn=nan

=n*2^(2n- 1)

sn= 1*2^ 1+2*2^3+3*2^5+4*2^7+5*2^9+......+n*2^(2n- 1) ( 1)

2^2*sn= 1*2^3+2*2^5+3*2^7+4*2^9+......+(n- 1)*2^(2n- 1)+n*2^(2n+ 1)(2)

( 1)-(2)=-3sn=2^ 1+2^3+2^5+2^7+2^9+......+2^(2n- 1)-n*2^(2n+ 1)

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

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

Sn = n * 2 (2n+1)/3-2 (4n-1)/9 can be obtained.

I hope the paper can help you. Welcome to ask questions.