在数列｛an｝中,a1=1,当n≥2时,n2=a1a2.an恒成立,则a3+a5等于?试比较an与a n+1的大小

a1a2...a(n-1)an=n² (1)
a1a2...a(n-1)=(n-1)² (2)
(1)/(2)
an=n²/(n-1)²
a3+a5=3²/(3-1)²+5²/(5-1)²=9/4+25/16=61/16
a(n+1)/an=[(n+1)²/n²]/[n²/(n-1)²]
=[(n+1)²(n-1)²]/n⁴
=[(n+1)(n-1)]²/n⁴
=(n²-1)²/n⁴a(n+1)