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.

Printable View

- Dec 22nd 2009, 11:40 PMPn0yS0ld13rFormula for nth term of sequence
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.

- Dec 23rd 2009, 01:15 AMTwistedOne151Hint
Here's a hint: consider the fact that $\displaystyle (2^k-1)[(2^k-1)+2]=(2^k-1)(2^k+1)=2^{2k}-1$.

--Kevin C. - Dec 23rd 2009, 02:02 AMShanks
$\displaystyle a_{n+1}=a_{n}\left(a_{n}+2\right)\text{ implies }a_{n+1}+1=a_{n}\left(a_{n}+2\right)+1,\text{ that is }a_{n+1}+1=\left(a_{n}+1\right)^2$

Construct a new sequence by letting $\displaystyle b_n=a_n+1$, you can take it from here. - Dec 23rd 2009, 08:09 AMSoroban
Hello, Pn0yS0ld13r!

I found the answer*by inspection*. . .

Quote:

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: .$\displaystyle a_n \;=\;2^{2^n}-1$