由题意 2an-1an+(n-1)an=3nan-1
故(n-1)/an-1+2=3n/an
设bn=n/an
则bn=bn-1/3 +2/3
故(bn-1)=1/3(bn-1 - 1)
bn-1为等比数列 又b1-1=-1/3
故bn=1-(1/3)^n
故an=n/(1-(1/3)^n)
a1*a2=1/(1-1/3)*2/(1-1/9)=3/2*9/4=27/8
由题意 2an-1an+(n-1)an=3nan-1
故(n-1)/an-1+2=3n/an
设bn=n/an
则bn=bn-1/3 +2/3
故(bn-1)=1/3(bn-1 - 1)
bn-1为等比数列 又b1-1=-1/3
故bn=1-(1/3)^n
故an=n/(1-(1/3)^n)
a1*a2=1/(1-1/3)*2/(1-1/9)=3/2*9/4=27/8