Question

数列问题！！！

在数列中 a1=1对于所有的n>=2都有a1 a2 a3 .....an=n^2求a3+a5

Answer 1

当n=2时
a1a2=4且a1=1
∴a2=4
当n=3时
a1a2a3=9
∴a3=9/4
当n=4时
a1a2a3a4=16
∴a4=16/9
当n=5时
a1a2a3a4a5=25
∴a5=25/16
∴a3+a5=(9/4)+(25/16)=61/16

Answer 2

a1*a2*a3*……*an
=a1*a2*a3*……*an(n-1)*an
=(n-1)^2*an
=n^2
an=(n^2)/(n-1)^2=[n/(n-1)]^2
a3=(3/2)^2=9/4
a5=(5/4)^2=25/16
a3+a5=9/4+25/16=61/16

Answer 3

an=n^2/(n-1)^2

Answer 4

a1a2=2^2=4
a1a2a3=3^=9

a3=9/4

a1a2a3a4=4^2=16
a1a2a3a4a5=5^2=25

a5=25/16

a3+a5=9/4+25/16=61/16

Answer 5

a1a2....an=n^2 1
a1a1....a(n-1)=(n-1)^2 2
1/2得
an=(n/n-1)^2
a3=9/4 a5=25/16
原=9/4+25/16=61/16