Open Question: what is the nth term of this sequence?
16:51 Publicado por Flechado16
1, 2/(1x3), 4/(1x3x5), 8/(1x3x5x7)...
=> 2^0/1, 2^1/(1x3), 2^2/(1x3x5), 2^3/(1x3x5x7)...
n th term = 2^(n-1) /[(2n-1)*(2n-3)*(2n-5)..........1]
numerator = 2^(n-1)
denominator = (2n-1)(2n-3)(2n-5)...........1
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