SOLUTION: Find the first five (5) terms of the sequence (begin with n = 1). A=n!/n^2 I still do not understand how to do this one.

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Question 23417: Find the first five (5) terms of the sequence (begin with n = 1).
A=n!/n^2
I still do not understand how to do this one.
Found 2 solutions by rapaljer, stanbon:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
First, hopefully you know what n! means.

For examples:
1! = 1
2! = 2*1 = 2
3!= 3*2*1=6
4!=4*3*2*1 = 24
5! = 5*4*3*2*1 = 120
etc.
n! = the product of the integers beginning with n and counting down to 1.

Now, A=+%28n%21%29%2F%28n%5E2%29+
Beginning with n=1:
n=1, so A+=+1%2F%281%5E2%29=+1
n=2, so A+=+%282%2A1%29%2F%282%5E2%29=+2%2F4=1%2F2
n=3, so A+=+%283%2A2%2A1%29%2F%283%5E2%29=+6%2F9=+2%2F3
n=4, so A+=+%284%2A3%2A2%2A1%29%2F%284%5E2%29=+24%2F16=3%2F2
n=5, so A+=+%285%2A4%2A3%2A2%2A1%29%2F%285%5E2%29=+120%2F25=+24%2F5

R^2 at SCC

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A1= 1!/1^2=1/1 = 1
A2= 2!/2^2= 2/4 = 1/2
A3= 3!/3^2= 6/9= 2/3
A4=4!/4^2= 24/16= 3/2
A5=5!/5^2=120/25= 24/5
Cheers,
Stan H.