*Posted by Mitchell in : potd , trackback
*
Given that the sequence $\{a_{n}\}_{n=0}^\infty$ satisfies $a_0 = 1$ and $a_{n+1} – a_n =2^n a_{n}^2$ for all $n \geq 0$, find a closed-form expression for $a_n$.

show

The expression \[a_n = \frac{3^{2^n} - 1}{2^{n+1}}\] is easily checked by induction.

Since I imagine this might not be a satisfactory explanation for some of you, I will explain how to arrive at this answer:

Multiply both sides of the equation by $2^{n+1}$. Then, we arrive at $b_{n+1} = 2 b_n^2 + 2 b_n$ for $n \geq 0$, where $b_n = 2^n a_n$.

This can be rearranged to the more simple-looking form $2 b_{n+1} + 1 = (2b_n + 1)^2$. This means that $2 b_n + 1 = (2 b_0 + 1)^{2^n}$ for all $n$. Since $b_0 = 2^0 a_0 = 1$, this means \[b_n = \frac{3^{2^n} - 1}{2}.\] Then, by the earlier relation $b_n = 2^n a_n$, we finally get \[\boxed{a_n = \frac{3^{2^n} - 1}{2^{n+1}}}.\]

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