0 votes 0 votes Solve the recurrence relation $a^{2}n-5a^{2}_{n-1}+4a^{2} _{n-2}=0$, if $a_{0}=4, a_{1}=13, n>1$ Combinatory discrete-mathematics combinatory recurrence-relation + – kidussss asked Jul 8, 2022 • recategorized Jul 8, 2022 by Arjun kidussss 489 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Nagend123 commented Jul 13, 2022 reply Follow Share @ankitgupta.1729. @kidussssIt is $a^2n$ and not $a^2_n$ ? 0 votes 0 votes ankitgupta.1729 commented Jul 14, 2022 reply Follow Share @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 $n \geq 0$ 0 votes 0 votes mahendrapatel commented Jul 15, 2022 reply Follow Share na,i think its a square index n 0 votes 0 votes Please log in or register to add a comment.