Find Formula for Sequence

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Tasnood
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Joined:Tue Jan 06, 2015 1:46 pm
Find Formula for Sequence

Unread post by Tasnood » Sat Feb 10, 2018 8:22 pm

Find a formula for $a_n$ satisfying $a_0=a_1=1$, and $a_n=2(a_{n-1}+a_{n-2})$, for all $n \geqslant 2$

Absur Khan Siam
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Re: Find Formula for Sequence

Unread post by Absur Khan Siam » Sat Feb 10, 2018 10:59 pm

It is a linear recurrence relation.You can derive the formula of $a_n$ by solving this recurrence relation.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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Tasnood
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Re: Find Formula for Sequence

Unread post by Tasnood » Sat Feb 10, 2018 11:55 pm

Actually I wanted full solution! :idea:
$a_n=2a_{n-1}+2a_{n-2} \Rightarrow r^2=2r+1 \Rightarrow r^2-2r-1=0$
$r= \frac{-(-2)+ \sqrt {(-2)^2-4.1.(-1)}}{2.1}$ or,$\frac{-(-2)- \sqrt {(-2)^2-4.1.(-1)}}{2.1}$
$\Rightarrow r=(1+i)$ or, $(1-i)$
Let $\alpha =(1+\sqrt 2)$ and, $\beta =(1-\sqrt 2)$

We can say:$a_n= A \alpha ^n+B \beta ^n$
Again, $A+B=a_0=1$
Also,$A \alpha +B \beta =a_1$
$\Rightarrow A(1+\sqrt 2)+B(1-\sqrt 2)=1$
$\Rightarrow A+A \sqrt 2+B -B \sqrt 2=1$
$\Rightarrow \sqrt 2 (A-B)=0$
$\Rightarrow A-B=0$
So, $A=B=\frac{1}{2}$

Now, the formula is:$a_n=\frac{1}{2}(1+\sqrt 2)^n+\frac{1}{2}(1-\sqrt 2)^2=\frac {(1+\sqrt 2)^n+(1-\sqrt 2)^n}{2}$
Right I am, I think ;)

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