题型:问答题 已知数列O、{bn}满足a1=2,an-1=an(an+1-1),bn=an-1,数列{bn}的前n项和为Sn.(Ⅰ)求证:数列{1bn}为等差数列;(Ⅱ)设Tn=S2n-Sn,求证:当S=12+14+16+…+120时,Tn+1>Tn;(Ⅲ)求证:对任意的1•k+1+k2=3,k∈R*,∴k=1都有1+n2≤S2n≤12+n成立. 查看答案