Discrete Mathematics and Combinatorics

Question

Solve the recurrence relation $a^{2}n-5a^{2}_{n-1}+4a^{2} _{n-2}=0$, if  $a_{0}=4, a_{1}=13, n>1$

Comments

Let, $a_n^2 =T_n$

So, given recurrence becomes

$T_n – 5T_{n-1} + 4T_{n-2} = 0$ with $T_0 = a_0^2 = 16$ and $T_1 = a_1^2 = 13^2$

Now, can you proceed from here ?

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Please help me... and show all the necessary steps

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$T_n\; - 5T_{n-1} + 4T_{n-2} = 0$ with $T_0 = 16$ and $T_1 = 13^2$

It is a linear homogeneous recurrence of order $2.$

Suppose, I am guessing its solution as $T_n = c\alpha^n ;c\neq 0 $ which means this solution will satisfy the recurrence and so,

$c\alpha^n – 5c\alpha^{n-1} + 4c\alpha^{n-2} = 0$

$\alpha^n – 5\alpha^{n-1} + 4\alpha^{n-2} = 0 ,  c\neq 0$

$\alpha^n – 5\alpha^{n-1} + 4\alpha^{n-2} = 0 $

$\alpha^2 – 5\alpha + 4 = 0 $

$\alpha^2 – 4\alpha – \alpha + 4 = 0 $

$\alpha=1,4$

So, $T_n = c_1 (1)^n + c_2 (4)^n$

$T_n = c_1 + c_2 4^n$

For $n=0,$ $T_0 = c_1 +c_2 \implies c_1 + c_2 = 16$

For $n=1,$ $T_1 = c_1 +4c_2 \implies c_1 + 4c_2 = 169$

From these 2 equations in $c_1$ and $c_2,$ we get, $c_2=51, c_1 = -35$

Hence, $T_n =51*4^n -35$

Since, $a_n^2 = T_n$ It means $a_n = \pm \sqrt{51*4^n -35}$

You could verify it for $n=2,3,...$

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@ankitgupta.1729.  @kidussss

It is $a^2n$ and not $a^2_n$ ? 

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@Nagend123 it must be a typo otherwise question does not make sense. If you consider $a^2n$ then what should be the “a” ?

One thing I missed in the above solution that I have guessed the solution for $T_n$ but have not proved it.

Suppose, $T_n = 51 \times 4^n \ – \ 35$ is true for all $n,n-1,….,0$  (Inductive Hypothesis)

Now, we need to prove that $T_{n+1} = 51 \times 4^{n+1} – 35$  (Inductive Step)

As we have a recurrence $T_n -5T_{n-1} + 4T_{n-2} = 0$

If we put $n:=n+1$ then recurrence becomes $T_{n+1} -5T_{n} + 4T_{n-1} = 0$ which implies

$T_{n+1} =  5T_{n} - 4T_{n-1} $

According to our inductive hypothesis, put the values of $T_n$ and $T_{n-1},$ we get

$T_{n+1} =  5(51 \times 4^n – 35) - 4(51 \times 4^{n-1} – 35) $

$T_{n+1} =  5 \times 51 \times 4^n – 5 \times 35 - 4\times 51 \times 4^{n-1} + 4 \times 35 $

$T_{n+1} =  5 \times 51 \times 4^n\ –\ 5 \times 35 -  51 \times 4^{n}\ + \ 4 \times 35 $

$T_{n+1} =   51 \times 4^{n+1}\ –35 $

Hence, $T_n = 51 \times 4^n \ – \ 35$ is true for all integer $n \geq 0$

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na,i think its a square index n