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n=1 a(1) = 0.001

n=2 a(2) = 0.012

n=3 a(3) = 0.144

n=4 a(4) = 1.728

n=5 a(5) = 20.736

....

n=k a(k) = (12^(k-1))/1000

let n = k+1

a(k+1) = 12^(k)/1000

12(ak) = a(k+1)

12(12^k-1)/1000 = 12^k/1000

the 12 gets absorbed here.

12^(k-1+1)/1000 = 12^k/1000

Valid for k and k+1

therefore our equation

A(n) = 12^(n-1)/1000, n(greater than or equal to) 1

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