在数列an中,a1=2,a2=4,an+1=3an-2an-1,设bn=log2(an+1-an)求证bn是等差数列,求数列1/bnbn+1的前n项和

an+1=3an-2an-1则a(n+1)-an=2(an-a(n-1))所以{a(n+1)-an}是以a2-a1=2是为首项,2为公比的等比数列所以a(n+1)-an=2*2^(n-1)=2^n而bn=log2(an+1-an)=log2(2^n)=n所以bn是等差数列,令cn=1/bnbn+1=1/[n(n+1)]=1/n-1/(n+1)...