Answer to Question #304666 in Discrete Mathematics for Tege

Question

11. For arbitrary constants c₁,c₂, And c₃ show that a_n=c₁2ⁿ+c₂5ⁿ+c₃n5ⁿ satisfies the recurrence relations a_n-12a_n-1+45a_n-2-50a_n-3=0

Answer 1

To prove $a_n=c_12^n+c_25^n+c_3n5^n$  satisfies the recurrence relation $a_n-12a_{n-1}+45a_{n-2}-50a_{n-3}=0$, it is enough to show that the solution of the recurrence relation is $a_n=c_12^n+c_25^n+c_3n5^n$.

The auxiliary equation of the recurrence relation is, $r^3-12r^2+45r - 50 =0.$

Using synthetic division,

$\begin{array}{l|lll} 2&1&-12&~~~45&-50\\ & 0 & ~~~~~2 & -20 & ~~~50\\ \hline & 1 & -10 & ~~~25 & \mid 0\\ \hline \end{array}$

We get 2 is a root of the equation. Hence,

$\begin{aligned} 0 = r^3-12r^2+45r - 50 &= (r-2)(r^2-10r+25)\\ &=(r-2)(r-5)^2 \end{aligned}$

Thus the roots of the auxiliary equation are, $r=2,5,5$.

The solution of the recurrence relation is

$a_{n} = c_12^n + (c_2 + c_3n)5^{n} = c_12^n + c_25^{n} + c_3n5^{n}$. Hence $a_{n}$ satisfies the given recurrence relation.