# `1, 3, 3^2/2, 3^3/6, 3^4/24, 3^5/120` Write an expression for the apparent nth term of the sequence. (assume that n begins with 1)

The numerator is obviously `3^(n-1),` the denominator is `(n-1)!`  (factorial).

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Let us first write the first two terms as fractions as well.

`1/1,3/1,3^2/2,3^3/6,3^4/24,3^5/120,...`

In the numerator we have powers of 3 (`3^0=1` and `3^1=3`). In the denominator we have factorials (`0! =1, 1! =1, 2! =2, 3! =6, 4! =24, 5! =120`). Therefore, the `n`th term of the sequence is

`a_n=3^(n-1)/((n-1)!)`

We have to put `n-1` because both powers and factorials start with 0 (`a_1=3^0/(0!)`)

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