Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Task is to find the explicit expression for the given recursive sequence with the help of power series.

Given: $a_{0}=0,\ a_{1}=1 \quad$ and $\quad a_{n}=5\cdot a_{n-1} -6\cdot a_{n-2}\quad $ for $\quad n \geq 2$

My Idea: despite the fact, i feel completly lost, about this problem, i thought i could be possible, to somehow build the formula for a power series-expression, then building the first derivative in order to find the explicit formula due to integrating.

finding $f(x) = a_{n}$ somehow like that:

$\int\limits_{0}^{x} \left( \sum\limits_{n=0}^{\infty}a_{n}\cdot x^{n} \right)^{\prime}\ =\ f(x)$

But i don't even know, if that is possible and how.. what do you think?

share|improve this question

3 Answers 3

up vote 4 down vote accepted

Use Wilf's "generatingfunctionology" techniques. Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write: $$ a_{n + 2} = 5 a_{n + 1} - 6 a_n $$ Multiplying by $z^n$ and add over $n \ge 0$. This gives: $$ \frac{A(z) - a_0 - a_1 z}{z^2} = 5 \frac{A(z) - a_0}{z} - 6 A(z) $$ Solve this for $A(z)$: $$ A(z) = \frac{z}{1 - 5 z + 6 z^2} = \frac{1}{1 - 3 z} - \frac{1}{1 - 2 z} $$ This is just two geometric series: $$ a_n = 3^n - 2^n $$

share|improve this answer

The characteristic equation is $$r^2-5r+6=0$$ and its roots are $r_1=2$ and $r_2=3$ hence $$a_n=\alpha (2)^n+\beta (3)^n$$ We find $\alpha$ and $\beta$ by $a_0$ and $a_1$.

share|improve this answer
A good solution, but not by the method of power series. –  GEdgar May 10 '13 at 14:19

Hint: For linear homogeneous recurrence relations with constant coefficients you can try a solution of the form $a_n=r^n$. If you put it into your recurrence, you get $r^2=5r-6$ This has two roots, $r_1,r_2$ and your general solution is $a_n=br_1^n+cr_2^n$. If you use your initial conditions you can evaluate $b,c$

share|improve this answer
A good solution, but not by the method of power series. –  GEdgar May 10 '13 at 14:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.