Find all sequences - recurrence relation

Question

Find all sequences $a_{n}$ whose members satisfy the recurrence relation $a_{n}=\frac{7}{10}a_{n-1}+\frac{3}{10}a_{n-2}+156n^{2}-280n+161$.

for $n ≥ 3$ (express the member $a_{n}$ as a function of its sequence number n).

I tried characteristic equation and partial solution, but partial solution is not working.

My solution: Characteristic equation: $a_{n}-\frac{7}{10}a_{n-1}-\frac{3}{10}a_{n-2}=0$ then substitution:

$r^2-\frac{7}{10}r-\frac{3}{10}=0$

So $r_{1}=1$ and $r_{2}=\frac{-3}{10}$

So characteristic equation is $c_{1}(1)^n+c_{2}(\frac{-3}{10})^n$

Then I tried partial solution when I used $an^2+bn+c$. But when I wanted to subtitute to sequence, then I didn't get good solution.

Answer 1

Observe that $\frac{7}{10} + \frac{3}{10} = 1$, we have \begin{align*} a_n - a_{n-1}= -\frac{3}{10}(a_{n-1}-a_{n-2})+156n^2-280n+161 \end{align*} Let $b_n = a_n - a_{n-1}$ for $n \geq 1$, the recurrence reduces to first order: \begin{equation*} b_n = -\frac{3}{10}b_{n-1}+156n^2-280n+161 \end{equation*} Once you obtain the solution to $b_n$, you should be able to find \begin{equation*} a_n = (\sum_{i=1}^{n} b_n) + a_0 \end{equation*}

Answer 2

The particular solution is cubic. I tried the general cubic equation but could not determine all coefficients. So instead we can use generating functions to solve the recurrence. Given the recurrence:

$$ a_n = \frac{7}{10} a_{n - 1} + \frac{3}{10} a_{n - 2} + 156n^2 - 280n + 161 $$

We convert to power series:

$$ \sum_{n = 2}^{\infty} a_n x^n = \frac{7}{10} \sum_{n = 1}^{\infty} a_n x^{n + 1} + \frac{3}{10} \sum_{n = 0}^{\infty} a_n x^{n + 2} + \sum_{n = 2}^{\infty} 156n^2 x^n + \sum_{n = 2}^{\infty} -280nx^n + \sum_{n = 2}^{\infty} 161 x^n $$

With $F(x) = \sum_{n = 0}^{\infty} a_n x^n$ we get:

$$ F(x) - a_0 - a_1x = \frac{7}{10} x \left(F(x) - a_0 \right) + \frac{3}{10} x^2 F(x) + \sum_{n = 2}^{\infty} 156n^2 x^n + \sum_{n = 2}^{\infty} -280nx^n + \sum_{n = 2}^{\infty} 161 x^n $$

Using rules of convergence:

$$ \sum_{n = 0}^{\infty} n^2 x^n = \frac{(1 + x)x}{(1 - x)^3} $$ $$ \sum_{n = 0}^{\infty} n x^n = \frac{x}{(1 - x)^2} $$ $$ \sum_{n = 0}^{\infty} x^n = \frac{1}{1 - x} $$

and rearranging gives us:

$$ F(x) = \frac{a_0 + (a_1 - 0.7a_0 - 37)x}{1 - 0.7x - 0.3x^2} + \frac{156x(x + 1)}{(1 - x)^3 (1 - 0.7x - 0.3x^2)} - \frac{280x}{(1 - x)^2(1 - 0.7x - 0.3x^2)} + \frac{161}{(1 - x)(1 - 0.7x - 0.3x^2)} - \frac{161}{1 - 0.7x - 0.3x^2} $$

Applying partial fraction decomposition and taking $a = a_0$ and $b = a_1 - 0.7a_0 - 37$:

$$ F(x) = \left( \frac{10}{13} \left( a - \frac{10}{3}b \right) - \frac{7700}{169} \right) \frac{1}{x + \frac{10}{3}} + \left( -\frac{10}{3}(a + b) + \frac{7700}{169} \right) \frac{1}{x - 1} + \frac{4410}{13(x - 1)^2} + \frac{520}{(x - 1)^3} + \frac{240}{(x - 1)^4}$$

$$ = \frac{\alpha}{x + \frac{10}{3}} + \frac{\beta}{x - 1} + \frac{4410}{13(x - 1)^2} + \frac{520}{(x - 1)^3} + \frac{240}{(x - 1)^4}$$

$$ = \frac{\alpha}{x + \frac{10}{3}} - \frac{\beta}{1 - x} + \frac{4410}{13(1 - x)^2} - \frac{520}{(1 - x)^3} + \frac{240}{(1 - x)^4}$$

Converting back to powerseries we find:

$$ a_n = \alpha \left( -\frac{3}{10} \right)^n - \beta + \frac{4410}{13}(n + 1) - 260(n + 2)(n + 1) + 40(n + 3)(n + 2)(n + 1)$$

which we can verify using induction.