How do I write the rule for this sequence that is neither arithmetic nor geometric as an equation? [closed]

$\begingroup$

I have the sequence $3, 5, 9, 17, 33, 65,\dots$

I know the rule is multiply by $2$ then subtract one. $3$ times $2$ is $6$ minus $1$ is $5$ and so on. Easy enough.

How do I write the rule for this sequence as an equation? I know it is neither arithmetic or geometric.

$\endgroup$ 4

4 Answers

$\begingroup$

You can write the sequence recursively as

$$x_0 = 3$$

$$x_n = 2x_{n-1} - 1 $$

To write it non-recursively:

Consider a new sequence

$$3, 6, 12, 24, 48, 96, ...$$

and your original:

$$3, 5, 9, 17, 33, 65, ...$$

Taking the difference of the two sequences we have

$$0, 1, 3, 7, 15, 31, ...$$

which is $g(x) = 2^{x-1} - 1$.

The top sequence is $3 * 2^{x-1}$. Subtracting $3 * 2^{x-1} - (2^{x-1} - 1)$.

Therefore, your function is

$$ f(x) = 2^x + 1$$

Also, note that you could also add $1$ to $2^x$ and get the same sequence.

$\endgroup$ $\begingroup$

You can write recursive function like this

$f(x)=\begin{cases} \ 2(f(x-1))-1 , & \text{if x>1,} \\[2ex] 3, & \text{if x=1} \end{cases}$

$\endgroup$ 2 $\begingroup$

We are given $$a_{n+1}=2a_n-1\tag1$$

Equivalently, $$a_{n+1}-2a_n=-1\tag2$$

and $$a_n-2a_{n-1}=-1.\tag3$$

Subtracting $(2)-(3)$ yields$$a_n-3a_n+2a_{n-1}=0\tag4$$

The characteristic equation is $r^2-3r+2=(r-1)(r-2)=0$.

Therefore, the solutions are $a_n=A+B\cdot2^n.$

Can you use the first two elements of the series to solve for $A$ and $B$?

$\endgroup$ 0 $\begingroup$

$$a_n=2a_{n-1}-1 \implies a_n - 1 = 2(a_{n-1}-1)$$Hence $a_n-1$ is a geometric sequence, and$$a_n-1 = 2^{n-1}(a_1-1) = 2^n \implies a_n = 2^n+1.\blacksquare$$

$\endgroup$