For n=1, we have s_1=a_1=t; for n=2, we find that a_2^2-2ta_2-3t^2=0. Since a_i's are all positive, so a_2=3t. Therefore, we guess than in general a_n=(2n-1)t and s_n=n^2 t. This can be easily proved by induction on n. (2)(\sqrt(2n+1)-\sqrt(2n-1))t\sqrt(n) =2t\sqrt(n)/(\sqrt(2n+1)+\sqrt(2n-1)) =\sqrt(2)t, when n approaches infinity. (3)n\sqrt(t)/(2n-1)t 1/2.
求什麽的？求t>0时Sn是什么数例还是…？
答：假设n个自然数是a1，a2，a3，…，an，而且考虑如下形式的和：S1=a1，S2=a1+a2，Sn=a1+a2+a3+…+an．如果在这n个和S1，S2，Sn...详情>>
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