Formula for nth term of sequence

Printable View

• Dec 22nd 2009, 11:40 PM
Pn0yS0ld13r
Formula 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 AM
TwistedOne151
Hint
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 AM
Shanks
$\displaystyle a_{n+1}=a_{n}\left(a_{n}+2\right)\text{ implies }a_{n+1}+1=a_{n}\left(a_{n}+2\right)+1,\text{ that is }a_{n+1}+1=\left(a_{n}+1\right)^2$
Construct a new sequence by letting $\displaystyle b_n=a_n+1$, you can take it from here.
• Dec 23rd 2009, 08:09 AM
Soroban
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.

I cranked out the first few terms . . .

. . $\displaystyle \begin{array}{|c|ccc|} n & & a_n & \\ \hline 1 & 3 &=& 2^2-1 \\ 2 & 15 &=& 2^4-1\\ 3 & 255 &=& 2^8-1\\ 4 & 65,\!535 &=& 2^{16}-1 \\ 5 & 4,\!294,\!967,\!295 &=& 2^{32}-1 \\ \vdots & \vdots & & \vdots \end{array}$

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