Basicaly, I need help on this sequence, as my homework is due in tomorrow. One question on sequences is:

1

3

11

67

I'm supposed to find out what comes next. Can anyone help? Even if you tell me how you got the answer, that would be a great help

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- Oct 8th 2007, 09:40 AMsnakeysterNeed help figuring out this sequence...
Basicaly, I need help on this sequence, as my homework is due in tomorrow. One question on sequences is:

1

3

11

67

I'm supposed to find out what comes next. Can anyone help? Even if you tell me how you got the answer, that would be a great help - Oct 8th 2007, 10:00 AMCaptainBlack
- Oct 8th 2007, 10:33 AMJhevon
i think $\displaystyle a_n = 7n^3 - 39n^2 + 70n - 37$ for $\displaystyle n \in \mathbb {N}$ works

thus, the fifth term is: 213

(i found the cubic that hit the points (1,1), (2,3), (3,11) and (4,67) by solving a system of equations) - Oct 8th 2007, 10:42 AMSoroban
Hello, snakeyster!

Quote:

I need help on this sequence: .$\displaystyle 1,\,3,\,11,\,67,\,\cdots$

. . $\displaystyle \begin{array}{ccccc} a_1 & = & 1 & & \\

a_2 & = & {\color{red}2}(a_1)\,{\color{red}+\,1} & = & 3 \\

a_3 & = & {\color{red}4}(a_2)\,{\color{red}-\,1} & = & 11 \\

a_4 & = & {\color{red}6}(a_3)\,{\color{red}+\,1} & = & 67\end{array}$

I bet the next one is:

. . $\displaystyle \begin{array}{ccccc} a_5 & = & {\color{red}8}(a_4)\,{\color{red}-\,1} & = & 535\end{array}$

The general term seems to be: .$\displaystyle a_n\;=\;2(n-1)(a_{n\text{-}1}) + (\text{-}1)^n$ . for $\displaystyle n \geq 2$

- Oct 8th 2007, 10:43 AMJhevon
- Oct 8th 2007, 01:23 PMCaptainBlack
You made no mistake, one can chose the next term to have an arbitary

value and find a polynomial that will take all the given values and our arbitary

next value.

Your cubic does give the first four terms, and so is as valid a rule for the

next term as any other, and I suspect it involves no more arbitary constants

than does Soroban's

RonL - Oct 8th 2007, 01:23 PMtopsquark
- Oct 8th 2007, 01:24 PMJhevon
- Oct 8th 2007, 01:52 PMSoroban
HellO!

RonL and Dan are, of course, absolutely correct.

A sequence of numbers is meaningless unless there is a stipulation,

. . a promise that that the sequence "continues in a similar manner",

. . that there is indeed a "reasonable" pattern to the terms.

I've posted the following before, but I enjoy showing it off.

**Find the next term of the sequence:**.$\displaystyle \bf{1,\,3,\,5,\,7,\,\cdots}$

Answer:**8**

Justification

I was using this function: .$\displaystyle f(n) \;=\;-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98x + 48\right) $

"That's unfair!" you say?

Okay, here's a better explanation . . .

The sequence: .$\displaystyle 1,\,3,\,5,\,7,\,8,\,\cdots$ is the sequence of natural numbers

. . whose English names contain the letter "e".

- Oct 8th 2007, 11:02 PMCaptainBlack
My favourite is: find the next term in 1, 2, 4, 8, 16.

The answer that I am thinking of is 31. See here in particular scroll down to sequence A000127.

RonL