an 4n2n2 Write the first five terms of the sequence Assume

a_n = 4n^2/(n+2)!

Write the first five terms of the sequence. (Assume n begins with 0.)

Solution

an = 4n^2/(n+2)!

n = 0 ; a0 = 0

n = 1 ; a1 = 4/(1+2)!= 4!/3! = 4

n = 2 ; a2 = 4(2)^2/(2+2)! = 16/4! = 16/24 = 2/3

n = 3 ; a3 = 4(3)^2/(3+2)! = 36/5! = 36/120 = 3/10

n = 4 ; a4 = 4(4)^2/(4+2)! = 64/6! = 64/720 = 8/90 = 4/45