log2(an)=n-1,an=2^(n-1)
bn=(3n-1)*an=(3n-1)*2^(n-1)
b1+b2+b3+.+bn=∑[3n·2^(n-1)](1到n)-∑[2^(n-1)](1到n)
=3∑[n·2^(n-1)](1到n)-2^n+1
设Sn=∑[n·2^(n-1)](1到n),则2Sn=∑[n·2^n](1到n)
Sn=2Sn-Sn=1·2^1+2·2^2+3·2^3+…+(n-1)·2^(n-1)+n·2^n
-1·2^0-2·2^1-3·2^2-…-n·2^(n-1)
=-[2^1+2^2+2^3+…+2^(n-1)]-1+n·2^n=-2^n+1-1+n·2^n=(n-1)·2^n
∴b1+b2+b3+.+bn=3(n-1)·2^n-2^n+1=(3n-2)·2^n+1