已知数列{an}的前n项和为sn,且sn=2n^2+n,n∈N*,数列{bn}满足an=4倍的以2为底（bn）的对数+3,n∈N*

an=4n-1 bn=2^(n-1)
tn=an·bn=(4n-1)2^(n-1) 2tn=(4n-1)2^n
t(n-1)=(4n-5)2^(n-2) 2 t(n-1)=(4n-5)2^(n-1)
t(n-2)=(4n-9)2^(n-3) 2 t(n-2)=(4n-9)2^(n-2)
......... .........
t2=7·2¹ 2t2=7·2²
t1=3·2º 2t1=3·2¹
错位相减得
Tn=(4n-1)2^n-4(2^(n-1)+2^(n-2)+2^(n-3)+......+2¹)-3·2º
=(4n-1)2^n-4·(2^n-2)-3·2º
=(4n-5)2^n+5

答案在图片中

(1) Sn= 2n^2+n (1) S(n-1) = 2(n-1)^2 + (n-1) (2) (1) -(2) an= 2n^2+n - 2(n-1)^2 - (n-1) = 4n-1 an=4log(2)bn+3 4n-1 = 4log(2)bn+3 log(2)bn = n-1 bn = 2^(n-1) (2) anbn = (4n-1)(2^(n-1)) = 4[n2^(n-1)] - 2^(n-1) consider [x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n [(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1) 1+2x+..+nx^(n-1) = [ (x-1)(n+1)x^n) -(x^(n+1)-1) ]/(x-1)^2 = [nx^(n+1)-(n+1)x^n+1]/(x-1)^2 put x=2 1+2(2)+3(2)^2+..+n(2)^(n-1) =n2^(n+1)- (n+1)2^n +1 summation anbn =summation {4[n2^(n-1)] - 2^(n-1)} = 4(n2^(n+1)- (n+1)2^n +1) - (2^n-1) =(4n).2^(n+1) - (4n+5).2^n +5

我怎么算出an=4n 1？