Originally Posted by

**angel.white** $\displaystyle A_0 = 1$

$\displaystyle A_n = 2n * A_{n-1}$

So do the first few:

$\displaystyle A_0 = (1)$

$\displaystyle A_1 = (1)*2(1)$

$\displaystyle A_2 = (1)*2(1)*2(2)$

$\displaystyle A_3 = (1)*2(1)*2(2)*2(3)$

$\displaystyle A_4 = (1)*2(1)*2(2)*2(3)*2(4)$

So we see that we can rewrite it like so:

$\displaystyle A_4 = (1)*2(1)*2(2)*2(3)*2(4) = (2*2*2*2)(4*3*2*1) = 2^44!$

So

$\displaystyle A_n = 2^nn!$

To prove it:

[tex]A_1 = 2(1)A_0 = 2(1)(1) = 2

$\displaystyle A_1 = 2^11!=2$

So our formula is true for n=1

$\displaystyle (1)*2(1)*2(2)*2(3)*...*2(n) = 2^nn!$

The next iteration will be 2(n+1) so multiply both sides by this:

$\displaystyle (1)*2(1)*2(2)*2(3)*...*2(n)*2(n+1) = 2^nn!2(n+1)$

$\displaystyle (1)*2(1)*2(2)*2(3)*...*2(n)*2(n+1) = (2*2^{n})[n!(n+1)]$

$\displaystyle (1)*2(1)*2(2)*2(3)*...*2(n)*2(n+1) = 2^{n+1}(n+1)! = A_{n+1}$

So by induction, it is true.

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Does it give you an $\displaystyle A_0$ or $\displaystyle A_1$?

Also, can you use parenthases? I'm interpreting this as $\displaystyle A_n = A_{n-1}+1+2^{n-1}$ But I am not sure this is correct.