a(n)*a(n+1)=a(n)-3a(n+1)+8 证明{a(n)+4/a(n)-2}是等比数列,并求{an}的通项公式

问题描述:

a(n)*a(n+1)=a(n)-3a(n+1)+8 证明{a(n)+4/a(n)-2}是等比数列,并求{an}的通项公式
a1=1

a(n)*a(n+1)=a(n)-3a(n+1)+8a(n+1)(an+3) = an +8a(n+1) = (an+8)/(an+3)The aux. equationx^2+2x-8=0x= -4 or 2a(n+1) = (an+8)/(an+3)a(n+1)-2= (an+8)/(an+3)-2= (-an+2)/(an+3)1/[a(n+1) -2] = (an+...