Suppose that $\displaystyle a_{1} = 3$ and $\displaystyle a_{n+1}=a_{n}\left(a_{n}+2\right)$. Find a formula for the nth term of this sequence.
Hello, Pn0yS0ld13r!
I found the answer by inspection . . .
I cranked out the first few terms . . .Suppose that $\displaystyle a_1 = 3$ and $\displaystyle a_{n+1}\:=\:a_n\left(a_n+2\right)$.
Find a formula for the $\displaystyle n^{th}$ term of this sequence.
. . $\displaystyle \begin{array}{|c|ccc|}
n & & a_n & \\ \hline
1 & 3 &=& 2^2-1 \\
2 & 15 &=& 2^4-1\\
3 & 255 &=& 2^8-1\\
4 & 65,\!535 &=& 2^{16}-1 \\
5 & 4,\!294,\!967,\!295 &=& 2^{32}-1 \\
\vdots & \vdots & & \vdots
\end{array}$
And saw that: .$\displaystyle a_n \;=\;2^{2^n}-1$