Question

已知数列｛an}中，an>0且Sn=1/2(an+n/an),求数列｛an}的通项公式

Answer 1

因为A1=S1
==>A1=1/2×(A1+1/A1)
==>A1=1

因为当n≥2时
有An=Sn-S[n-1]

所以Sn=1/2×[(Sn-S[n-1])+n/(Sn-S[n-1])]

==>2Sn=(Sn-S[n-1])+n/(Sn-S[n-1])

==>2Sn(Sn-S[n-1])=(Sn-S[n-1])²+n

==>2Sn²-2SnS[n-1]=Sn²-2SnS[n-1]+S[n-1]²+n

==>Sn²-S[n-1]²=n

则有S[n-1]²-S[n-2]²=n-1

S[n-2]²-S[n-3]²=n-2

.....

S3²-S2²=3

S2²-S1²=2

将以上式子叠加可得Sn²-S1²=2+3+....+n

==>Sn²=1+2+3+...+n=n(n+1)/2

因为An>0
==>Sn>0

所以Sn=√n(n+1)/2

所以An=Sn-S[n-1]=√n(n+1)/2-√n(n-1)/2

经过检验An=√n(n+1)/2-√n(n-1)/2满足A1=1

所以An=√n(n+1)/2-√n(n-1)/2