# Question about converting an arithmetric sequence formula to an explicit geometric formula

I have been given the following sequence, and told to work out the formula: $$3,15,63,255,1023,\dots$$ I noticed that each term was being multiplied by 4 and then 3 was added to get the next term. From this trend, I deduced a recursive formula of: $$T_n=4T_{n-1}+3, T_1=3$$ From common sense, I then realised that an alternative way of looking at it was that the $n$th term was equal to $4^n-1$. That is: $$T_n=4^n-1$$ My question: is there a way that I can convert the recursive sequence to the geometric sequence? My attempt is shown below:

1. I deduced from the recursive formula that $a$ (the first term) was 3.
2. I looked only at the section of the recursive formula that looked geometric, i.e. $4T_{n-1}$.
3. Thus I used 4 as my common ratio $r$.
4. I subbed these values into the geometric sequence formula $T_n=ar^{n-1}$, like so: $$T_n=3\cdot4^{n-1}$$
5. I then added the plus 3 to this geometric sequence to get: $$T_n=3\cdot4^{n-1}+3$$

I know this is incorrect as I tested it out on the sequence and it didn't work.

## 1 Answer

If I have a linear recurrence $$x_n=ax_{n-1}+b$$ with given $x_1$ I can rewrite it as $$y_n=ay_{n-1}$$ with $y_n=x_n+\frac b{a-1}$. (If $a=1$ the recurrence is an arithmetic progression and has a trivial closed form). It is easy to see that $y_n$ follows a geometric progression: $$y_n=a^{n-1}y_1=a^{n-1}\left(x_1+\frac b{a-1}\right)$$ which leads to the final solution for $x_n$ as $$a^{n-1}\left(x_1+\frac b{a-1}\right)-\frac b{a-1}$$ $$=a^{n-1}x_1 + \frac b{a-1}(a^{n-1}-1)$$ For your case, $a=4$, $b=3$ and $T_1=3$, so the formula evaluates to $$T_n=3(4^{n-1}) + \frac3{4-1}(4^{n-1}-1)$$ $$=4^n-1.$$ Any linear recurrence can be solved in a similar way; see Wikipedia for more details.