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

4 Comments

It is $a^2n$ and not $a^2_n$ ?

@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 $n \geq 0$

na,i think its a square index n

0 votes
0 answers
1
229 views
10 votes
1 answer
2
469 views
1 vote
3 answers
3
434 views