# what is the pattern?

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• Apr 18th 2009, 02:40 PM
nithujaa
what is the pattern?
t1=0, find the first few terms tn=1/tn-1+2, what is the pattern?
• Apr 18th 2009, 03:20 PM
Soroban
Hello, nithujaa!

Quote:

Given: .$\displaystyle t_1=0,\;\;t_n\:=\:\frac{1}{t_{n-1}+2}$

Find the first few terms. .What is the pattern?

$\displaystyle \begin{array}{ccccccc} t_1 &=& 0 \\ t_2 &=& \frac{1}{0+2} &=& \frac{1}{2} \\ \\[-4mm] t_3 &=& \frac{1}{\frac{1}{2} + 2} &=& \frac{1}{\frac{5}{2}} &=& \frac{2}{5} \\ \\[-4mm] t_4 &=& \frac{1}{\frac{2}{5}+2} &=& \frac{1}{\frac{12}{5}} &=& \frac{5}{12} \\ \\[-4mm] t_5 &=& \frac{1}{\frac{5}{12}+2} &=& \frac{1}{\frac{29}{12}} &=& \frac{12}{29} \end{array}$

The fractions are: .$\displaystyle \frac{0}{1},\; \frac{1}{2}, \;\frac{2}{5},\; \frac{5}{12},\; \frac{12}{29},\; \frac{29}{70},\; \frac{70}{169},\; \frac{169}{408},\; \hdots$

The numerators form the sequence: .$\displaystyle 0,1,2,5,12,29, 70, 169, 408, \hdots$
The denominators are "one step ahead."

For the numerators, we have: .$\displaystyle a_1 = 0,\;a_2 = 1,\;a_n \:=\:2a_{n-1} + a_{n-2}$
. . Each term is twice the preceding term, plus the second preceding term.

If you require a "closed form" for the numerators, here it is:

. . $\displaystyle a_n \;=\;\frac{(1+\sqrt{2})^{n-1} - (1-\sqrt{2})^{n-1}}{2\sqrt{2}}$

• Apr 18th 2009, 03:39 PM
nithujaa
thx..but this question is for data management?..so can i use this method?