Question

(1/an)-an=2根号n,且an>0 (1)求数列｛an｝的通项公式,(2)证明a1+a2+……+an＜根号n

Answer 1

(1).(1/an)-an=2√n
[1-(an)^2]/an=2√n
1-(an)^2=(2√n)an
(an)^2 + (2√n)an-1=0
(an)^2 + (2√n)an +n -n-1=0
(an+√n)^2=n+1
an+√n=±√n+1
∵an>0
∴an=√n+1 - √n
(2)a1+a2+a3+.+an=(√2 -1) +(√3 -√2)+(2+√3)+.+(√n+1 - √n)=-1+√n+1=Sn
√n=(1/2)[(1/an)-an]
1.当n=1时,-1+√2＜1
2.假设当n=k时,结论成立,即ak=√k+1 -√k ,这时Sk=-1+√k+1 ,√k=(1/2)[(1/ak)-ak]
则 Sk+1 -√k+1 =(Sk +ak+1)-√k+1
=[(-1+√k+1)+(√k+2 -√k+1)]-(1/2)[(1/ak+1)-ak+1]
=(-1+√k+2)-(1/2){[1/(√k+2 -√k+1)] +(√k+2 -√k+1)}
=(-1+√k+2)-(√k+2)
=-1＜0
∴Sk+1＜√k+1
即：当n=k+1时,结论也成立
∴根据1,2知,a1+a2+.+an＜√n