已知数列{an}中,n属于N*,an>0 其前n项和为Sn 满足2根号下Sn=an+1
问题描述:
已知数列{an}中,n属于N*,an>0 其前n项和为Sn 满足2根号下Sn=an+1
求证an是等差数列
可以说清楚一点吗?
答
因为2√S(n) =a(n)+1 2√S(n+1)=a(n+1)+1所以两式平方相减4(S(n+1)-S(n))=[a(n+1)+1]^2-[a(n)+1 ]^24·a(n+1)=[a(n+1)]^2+2·a(n+1)-[a(n)]^2-2·a(n)[a(n+1)+a(n)]·[a(n+1)-a(n...