0
$\begingroup$

Given the sequence $6,15,18,19,...,$ what is the recursive/explicit formula?

This is clearly not an arithmetic or geometric series but it does seem to satisfy:

\begin{align*} u(1) &= 6 \\ u(n) &= u(n-1) + 9^{1/n} \end{align*}

A general question: Given only a few terms I would guess that my solution is not unique. Is there any way to prove this other than by counter-example?

Are there any good strategies for determining these formulae other than intuition and trial&error?

Thank you, Chris

$\endgroup$
  • $\begingroup$ For questions like these (where they give you the first few terms), you really can only use intuition. You could find a polynomial $p(x)$ which matches any finite sequence, i.e. $p(0)=a_0$, $p(1)=a_1$, etc. But that's usually more laborious than noticing a simple pattern. $\endgroup$ – Trevor Norton Dec 9 '15 at 15:48
  • $\begingroup$ Also, I don't think your answer is correct. For instance $9^{1/3}\neq 1$. And your indexing is off by one (you probably meant $u_0=6$). $\endgroup$ – Trevor Norton Dec 9 '15 at 15:52
  • $\begingroup$ oh I see, sloppy me: u(0) = 6, u(n) = u(n-1) + 9/(3^(n-1)). Thanks. $\endgroup$ – Chris Crawford Dec 9 '15 at 22:29
1
$\begingroup$

A sequence is just an ordered list of things; there doesn't necessarily have to be a single "nice" formula that goes with it. To show that your solution is not unique, just start listing other numbers. If we trust your recursive formula for the first few terms, the sequence could be $$ 6,15,18,19,0,0,0,0,0,\dotsc $$ given by the rule $$\begin{cases} u(1) = 6 \\ u(n) = u(n-1) + 9^{1/n} \quad\forall x \in \{2,3,4\}\\ u(n) = 0 \quad\text{otherwise} \end{cases}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think that sequence is actually right though. I made a comment under the main thread. $\endgroup$ – Trevor Norton Dec 9 '15 at 15:55
0
$\begingroup$

Could it be possible to have this sequence?

$$a_n = -\frac{3\cdot(n^2 - 22n + 15)}{2n+1}$$

But then you have to admit the possibility to have real numbers in the whole sequence.

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

Note that is $+9,+3,+1$. You can complete as you want... For example $+9,+3,+1$ recursively, or $a_n=a_{n-1}+3^{4-n}$, etc

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ totally wrong. 6,33,15,9,.. $\endgroup$ – RE60K Dec 9 '15 at 15:49
  • $\begingroup$ exponents are off - should be 3^(3-n) $\endgroup$ – Chris Crawford Dec 9 '15 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.