由Sn=a1+3?a2+32?a3+…+3n-1?an，所以3Sn=3a1+32?a2+33?a3+…+3n?an，相加4Sn=a1+3（a1+a2）+…+3n-1?（an-1+an）+3n?an，所以4Sn-3nan=1+3( 1 3 )1+32( 1 3 )2+…+3n-1?( 1 3 )n-1=n．故答案为n．
