数列an,满足a1=4,an+1*an+6an+1-4an-8=0,记bn=6/an-2,n属于正整数,
问题描述:
数列an,满足a1=4,an+1*an+6an+1-4an-8=0,记bn=6/an-2,n属于正整数,
1.求数列bn的通项公式 2.求数列{an*bn}的通项公式
答
1)由an=6/bn+2,代入等式得:[6/b(n+1)+2]*[6/bn+2]+36/b(n+1)+12-24/bn-8-8=036/b(n+1)bn+12/b(n+1)+12/bn+4+36/b(n+1)-24/bn-4=03/b(n+1)bn+4/b(n+1)-1/bn=0b(n+1)=3+4bnb(n+1)+1=4(bn+1)故{bn+1}为首项b1+1=6/(a1...