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How do you write a recursive formula for the sequence 1, 2, 3/2, 4/6, 5/24...?

1 Answer

Feb 23, 2016

$a_{1} = 1$ $a_1 = 1$

$a_{n + 1} = \frac{n + 1}{n^{2}} \cdot a_{n}$ $a_(n+1) = (n+1)/n^2 * a_n$

Explanation:

A direct formula would be:

$a_{n} = \frac{n}{(n - 1)!}$ $a_n = n/((n-1)!)$

So we find:

$\frac{a_{n + 1}}{a_{n}} = (\frac{n + 1}{n!}) \div (\frac{n}{(n - 1)!}) = \frac{(n + 1) (n - 1)!}{n \cdot n!} = \frac{n + 1}{n^{2}}$ $a_(n+1)/a_n = ((n+1)/(n!)) -: (n/((n-1)!)) = ((n+1)(n-1)!) / (n*n!) = (n+1)/(n^2)$

Hence:

$a_{n + 1} = \frac{n + 1}{n^{2}} \cdot a_{n}$ $a_(n+1) = (n+1)/n^2 * a_n$

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