设数列{an}满足a1=2,an+1-an=3*2^2n-1

(1)a2-a1=3*2^(2*1-1)
a3-a2=3*2^(2*2-1)
.
an-a(n-1)=3*2^(2*(n-1)-1)
上式累加得an-a1=3*[2^1+2^3+...2^(2*n-3)]=3*{[4*2^(2*n-3)-2]/(4-1)}=2^(2*n-1)-2
故an=2*4^(n-1)=2^(2*n-1)
(2)bn=n*2^(2n-1)
Sn = 1* 2^1 + 2*2^3 + 3* 2^5 +.+ n*2^(2n-1)
4*Sn = 1* 2^3 + 2*2^5 +.+(n-1)2^(2n-1) + n*2^(2n+1)
两式相减得
3*Sn=n*2^(2n+1)-2-[2^3+2^5.+2^(2n-1)]=n*2^(2n+1)-[2^(2*n-1)-2]/3
Sn=n*2^(2n+1)/3 - 2^(2n+1)/9 + 2/9