数列{an}满足递推式an=3a(n-1)+3^n-1(n>=2),又a1=5使得{(an+y)/3^n}为等差数列的实数y=

设bn=(an+y)/3^n要使其为等差数列,则bn-b(n-1)为一个常数bn-b(n-1)=(an+y)/3^n-[a(n-1)+y]/3^(n-1)然后把an=3a(n-1)+3^n-1代入求得bn-b(n-1)=1-(1+2y)/3^ny是实数,不能是关于n的代数式,故1+2y=0y=-1/2...