20th term in sequence

• March 8th 2011, 07:35 AM
mathguy80
20th term in sequence
Find the next and 20th term in the sequence.

1, -3, 2, -5, 4, -7, 8, -9...

This one seems very hard to me. I tried the difference table method, which gives no pattern that I can see.

1 -3 2 -5 4 -7 8 -9
-4 5 -7 9 -11 15 -17
9 -12 16 -20 26 -32
-21 +28 -36 46 58

Only thing I can think of so far is the alternate +ve and -ve values could be solved with (-1)^n, so it gives a negative value for odd powers.

This problem does not seem to be a simple arithmetic or geometric sequence. Can anyone guide towards what area of mathematics this problem is from, and is there a way to think of similar sequence problems?

Thanks.
• March 8th 2011, 07:41 AM
e^(i*pi)
Your odd terms in the sequence are doubling from the previous one

Your even terms form an arithmetic sequence with $a=-3$ and $d=-2$.

Can you find the next (odd) term and the 20th (even) term given the above?

edit: the question does not ask for the nth term so don't worry about it!
• March 8th 2011, 07:50 AM
Plato
Quote:

Originally Posted by mathguy80
Find the next and 20th term in the sequence.
1, -3, 2, -5, 4, -7, 8, -9...

Try $a_n = \left\{ {\begin{array}{rl}
{2^{\frac{{n - 1}}
{2}} ,} & {\text{if n is odd}} \\
{ - n - 1,} & {\text{if n is even}} \\ \end{array} } \right.$
• March 8th 2011, 08:02 AM
mathguy80
Didn't think of it that way at all! So its both an arithmetic sequence for the common difference in even terms and geometric for the odd terms?

So for odd terms its 2^(n-1) and for even terms it's -2n-1. So depending on if the odd/even term is required use corresponding function.

Is this correct, Having 2 different functions for the same sequence?
• March 8th 2011, 08:14 AM
TheChaz
Techincally, it's just one function, where even and odd terms are defined separately. This is certainly still a (well-defined) function.

Suffice it to say "yes", this is correct!
• March 8th 2011, 08:17 AM
Plato
Quote:

Originally Posted by mathguy80
So for odd terms its 2^(n-1) and for even terms it's -2n-1. So depending on if the odd/even term is required use corresponding function.

That does not work.
When $n=3$ it is odd.
BUT $2^{3-1}=4$, and that is not the third term.

It is quite common to use two or more expressions to define sequences.
• March 8th 2011, 08:25 AM
Soroban
Hello, mathguy80!

Quote:

$\text{Find the next and 20th term in the sequence: }\;1, \text{-}3, 2, \text{-}5, 4, \text{-}7, 8, \text{-}9 \hdots$

$e^{1\pi}$ is correct.

There are two sequences which have been riffle-shuffled together.

. $\begin{array}{c||c|c|c|c|c|c|c|c}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \hline
\text{odd} & 1 && 2 && 4 && 8 & \\ \hline
\text{even} && \text{-}3 && \text{-}5 && \text{-}7 && \text{-}9 \\ \hline \end{array}$

And Plato's piecewise function is perfect.

And now for my two cents . . .
The function can be written: . $a_n \;=\;\dfrac{1\!-\!(\text{-}1)^n}{2}\cdot 2^{\frac{n-1}{2}} - \dfrac{1\!+\!(\text{-}1)^n}{2}\cdot(n+1)$