A1=2,A(n+1)=2/(An+1),Bn=|(An+2)/(An-1)|,n=1,2,3...求数列Bn通项?
问题描述：
A1=2,A(n+1)=2/(An+1),Bn=|(An+2)/(An-1)|,n=1,2,3...求数列Bn通项?
已知数列{An}前n项和Sn,满足Sn=(9/8)An-(1/2)*3^n+0.5,n=1,2,3.
(1)求An通项；
(2)设Bn=(3^n)/(Sn),Tn=B1+B2+B3+...+Bn,求证：对于任意n=1,2,3...都有Tn<0.5.有看不懂的可以问）
答
(i)a(n+1)=2/(an+1)a(n+1) -1 =2/(an+1) -1= -(an-1)/(an+1)1/[a(n+1) -1] = -(an+1)/(an-1)= -1 - 2/(an-1)1/[a(n+1) -1]+1/3 = -2[ 1/(an-1) +1/3] => {1/(an-1) +1/3} 是等比数列,q=-21/(an-1) +1/3 = (-2)^(n-1)...