I'm trying to do the following exercise:
Find a non-homogeneous recurrence relation for the sequence whose general term is
$$a_n = \frac{1}{2}3^n – \frac{2}{5} 7^n$$
From this expression we can obtain the roots of the characteristic polynomial $P(x)$, which are $3$ and $7$, so $P(x) = x^2 – 10x + 21$ and $a_n = 10a_{n-1} – 21a_{n-2} \; \forall \; n \ge 2, \; a_0 = \frac {1}{10}, \; a_1 = -\frac{13}{10}$.
Now I don't know how to obtain a non-homogeneous recurrence relation given this homogeneous recurrence relation.
Best Answer
Write the relations for two consecutive terms:
$$ \begin{align} \begin{cases} a_n &=\, \dfrac{1}{2}\,3^n - \dfrac{2}{5}\, 7^n \\ a_{n+1} &=\, \dfrac{1}{2}\,3^{n+1} - \dfrac{2}{5}\, 7^{n+1} \,=\, \dfrac{3}{2}\,3^n-\dfrac{14}{5}\,7^n \end{cases} \end{align} $$
Eliminate (for example) $\,7^n\,$ between the two:
$$ \require{cancel} a_{n+1} - 7 a_n = \left(\dfrac{3}{2}\,3^n-\cancel{\dfrac{14}{5}\,7^n}\right) - \left(\dfrac{7}{2}\,3^n - \cancel{\dfrac{14}{5}\, 7^n}\right) \;\iff\; a_{n+1} = 7a_n-2\cdot 3^n $$
Note that the non-homogeneous recurrence is not unique. If you chose to eliminate the other power $\,3^n\,$, for example, you would get $\,a_{n+1} = 3a_n-\dfrac{8}{5}\,7^n\,$, which is equally valid, as are many others.