Question

数列an的首项均为正数,a1=1,且满足(n+2)an+1^2-nan^2+2an+1an=0 着它的前n项和sn=

Answer 1

解：
(n+2)a(n+1)²-nan²+2a(n+1)an=0
等式两边同除以an²
(n+2)[a(n+1)/an]² +2[a(n+1)/an] -n=0
[a(n+1)/an +1][(n+2)a(n+1)/an -n]=0
a(n+1)/an=-1(数列各项均为正，比值为正，舍去)或(n+2)a(n+1)/an=n
a(n+1)/an=n/(n+2)
an/a(n-1)=(n-1)/(n+1)
a(n-1)/a(n-2)=(n-2)/n
…………
a2/a1=1/3
连乘
an/a1=(1/3)(2/4)...[(n-1)/(n+1)]=[1×2×...×(n-1)]/[3×4×...×(n+1)]
=2/[n(n+1)]
an=2a1/[n(n+1)]=2/[n(n+1)]=2[1/n -1/(n+1)]
Sn=a1+a2+...+an
=2[1/1-1/2+1/2-1/3+...+1/n -1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)

Answer 2

苏轼