I was given this problem on a test:
Find an equation for the terms in this series: 1, 5, 29, 173, 1037
Now the only way I can even think to do is either common difference, or common ratio.
Trying common difference shows the next pattern to be 4, 24, 144, 864. So obviously the sub-pattern is represented by $4*6^{n-1}$. Now how do I get a general equation for the entire series from that?
I thought maybe I could plug in the equation for the sub series into $d$ in the equation:
$a + (n-1)*d$,
since that's the standard equation for an arithmetic series, but of course that doesn't work because the arithmetic series equation always references the first term.
No clue how to do this.
Ideally I just want to know if there's a general rule for an equation when you have to go multiple levels deep into common difference or common ratio, or both.
For example a harder one is: 3, 5, 13, 85, 3613. So common difference (CD) gives: 2, 8, 72, 3528. Then the next pattern if I used common ratio (CR) would be, 4, 9, 49, … . Then the next pattern from CD would be 4, 40, …
Is it even meaningful to alternate between CD and CR in this manner?
Best Answer
$$ \frac{4 \cdot 6^{n-1}+1}{5}$$
The general rule is called the Calculus of finite differences. We have \begin{gather*} b_1=a_2-a_1,\\ b_2=a_3-a_2,\\ \ldots \\ b_{n-1}=a_n-a_{n-1}. \end{gather*} Then $b_1+b_2+ \cdots +b_{n-1}=a_n-a_1.$ $$ \sum_{i=1}^{n-1} b_i=\sum_{i=1}^{n-1}4\,{6}^{n-1}=4 \sum_{i=1}^{n-1}{6}^{n-1}=4 \cdot \frac{6^n-6}{30}. $$ Then $$ a_n=4 \cdot \frac{6^n-6}{30}+1=\frac{4 \cdot 6^{n-1}+1}{5} $$