Question:
The first three terms of a sequence are given.
$2, 4, 8...$
Write two different rules for continuing the sequence. Give the next two terms for each rule.
Answer:
I have found one rule which satisfies this sequence and could not think of another one.
$1)$ The next term is double the previous term:
$2, 4, 8, 16, 32...$
I'm not entirely what another different rule could be... Thank you.
$\endgroup$ 45 Answers
$\begingroup$$f(n) = n^2 - n + 2$ gives $f(1) = 2, f(2) = 4, f(3) = 8, f(4) = 14, f(5) = 22$
You have $3$ terms so you can construct a $2$nd degree polynomial passing through those three points. There are two approaches to this, write $f(n) = an^2 + bn + c$ and then plug in $(1, 2), (2, 4), (3, 8)$ and find the coefficients which isn't too bad to do. On the other hand, you can use Lagrange Interpolation for a quick way to compute the polynomial.
$\endgroup$ 0 $\begingroup$One idea is to look at the differences between consecutive terms. $4-2=2$ and $8-4=4$, an increase of $2$, so we could try making these differences increase by $2$ each time. The next difference would have to be $4+2=6$, and the one after that would then be $6+2=8$; the corresponding terms of the sequence would be $8+6=14$ and $14+8=22$. Can you work out a formula for the $n$-th term if this rule is followed? The following table may help you do this.
$$\begin{array}{rcc} n:&1&2&3&4&5\\ x_n:&2&4&8&14&22\\ n^2:&1&4&9&16&25\\ n^2-x_n:&-1&0&1&2&3 \end{array}$$
$\endgroup$ $\begingroup$Here's a family of uncountably many sequences:
$f_\epsilon(n) = \lfloor (2+\epsilon)^n \rfloor$ with $0\le\epsilon<\sqrt[3]{9}-2$
$\lfloor x\rfloor$ denotes the largest integer not larger than $x$.
Here's another interesting sequence:
$f(n)$ is the sum of the digits of the $(n+4)$-th prime.
Or, $f(n)$ is the maximal number of pieces of a cake achievable with $n$ cuts.
Of course there's a ton of other possibilities.
$\endgroup$ $\begingroup$"2, 4, 8" is not a "sequence", it is the beginning of one- and, given any first three values, there are an infinite number of sequences that start with those three values.
$\endgroup$ 1 $\begingroup$Another possible rule is that the sequence is the sequence of even numbers not divisible by $3$. So $f(n) = 2n$ where $n \neq 0 \pmod 3$ that gives $f(n) = 2n$ where $n = 1, 2, 4, 5, 7, 8, \ldots$ this gives $f(1) = 2, f(2) = 4, f(4) = 8, f(5) = 10, f(7) = 14, f(8) = 16$.
Point being, you can think up all sorts of contrived rules that fit the first $k$ terms of a given sequence, i.e: listing the first few terms of a sequence doesn't uniquely identify the sequence.
$\endgroup$
