Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$

Question

Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$

I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.

Any help is appreciated! Thanks.

Edit: solution I require is $\frac{1}{2} (3^{n-1}+1-2^n)$

Edit2: Solutions to the homogeneous equation would be of the form $h_n = A(\alpha)^n + B(\beta)^n$, and $p_n$ will exist such that $s_n = h_n + p_n$

@Mhenni My problem is I do not know the correct "guess" for $h_n$. If I know that, I am equipped to do the rest (the links you showed me aren't helpful in that regard). Uros, yes, I'm sure. — Clinton, May 18 '13 at 08:34
haha somabha, i was given the answers at the back of the book, how do i use induction otherwise? :P — Clinton, May 18 '13 at 08:40
You can't use induction if you don't have a formula. That is why using induction is not well suited to formulate solutions but to test them. You need to use other methods in general to solve recurrence relations, or else the whole topic would be trivial as most mathematicians are well versed in enough induction to check solutions. — Jon Claus, May 18 '13 at 08:46

score 1 · Accepted Answer · answered May 18 '13 at 08:37

1

Let $ t_n := \frac{s_n}{3^n} $. Hence, $ t_n = t_{n-1} + 3^{-n}\left(2^{n-2} - 1\right) $. Also, $ t_1 = 0 $. Hence, $$ t_n = \sum_{k = 2}^n \frac{1}{4}\left(\frac{2}{3}\right)^n - 3^{-n} $$ This is a geometric series and easily evaluated to arrive at $$ t_n = \frac{1}{2}\cdot 3^{-n} \cdot \left(1 - 2^{n}\right) + \frac{1}{6} \implies s_n = \frac{1 + 3^{n-1}- 2^{n}}{2} $$

answered May 18 '13 at 08:37

Jon Claus

2,760

thank you! but i require to do it the substitution guessing method :\ – Clinton May 18 '13 at 08:39
The amount of notation you introduced without any sort of definitions makes it impossible to follow what you attempted. For one, you define $ p_n $ then don't even bother to use it. So I can't really tell where you went wrong or what you even tried. I did perform a substitution though, so it seems like you could refactor my solution into something you want. – Jon Claus May 18 '13 at 08:45
added what they meant - maybe it's a unique method to our lecturer :P – Clinton May 18 '13 at 08:52
never mind, got it, thanks! – Clinton May 18 '13 at 08:59

score 1 · Answer 2 · answered May 21 '13 at 00:44

A general technique is taught by Wilf's "generatingfunctionology". Define $S(z) = \sum_{n \ge 0} s_n z^n$ and write ($s_0$ you get from the recurrence "backwards", mostly for not having to mess around with indices): $$ s_{n + 1} = 3 s_n + 2^{n - 1} - 1 \qquad s_0 = \frac{1}{6} $$ Multiply the recurrence by $z^n$, add over $n \ge 0$ to get: $$ \begin{align*} \frac{S(z) - s_0}{z} &= 3 S(z) + \frac{1}{2} \sum_{n \ge 0} 2^n z^n - \sum_{n \ge 0} z^n \\ &= 3 S(z) + \frac{1}{2} \frac{1}{1 - 2 z} - \frac{1}{1 - z} \end{align*} $$ Solving for $S(z)$ and expanding in partial fractions: $$ S(z) = \frac{1}{6} \frac{1}{1 - 3 z} - \frac{1}{2} \frac{1}{1 - 2 z} + \frac{1}{2} \frac{1}{1 - z} $$ Everything in sight is geometric series: $$ s_n = \frac{1}{6} \cdot 3^n - \frac{1}{2} \cdot 2^n + \frac{1}{2} = \frac{1}{2} \left( 3^{n - 1} - 2^{n - 1} + 1 \right) $$

Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$

2 Answers2