# Thread: Finding the nth term of the following sequence

1. ## Finding the nth term of the following sequence

1) - 1/2 , 1/4 , - 1/2 , 1/4

2) 11, 8, 13, 6, 15,...

3) 8,4,8,4,8,4,...

and could you guys share any tips on how to quickly find the nth term.
any help is much appreciated. thank you

2. ## Re: Finding the nth term of the following sequence

Hello, nisbahmumtaz!

(1) and (3) are oscillating sequences.
There are several ways to express the general term.
I'll show you one of them.

$(1)\;\text{-}\tfrac{1}{2},\;\tfrac{1}{4},\;\text{-}\tfrac{1}{2},\;\tfrac{1}{4},\;\hdots$
Consider the expression: . $b(n) \:=\:\frac{1 + (\text{-}1)^n}{2}$

$\text{When }n\text{ is odd, }b(n) = 0.$
$\text{When }n\text{ is even, }b(n) = 1.$
. . I call it the "blinker" function.

$\text{Therefore, the }n^{th}\text{ term is: }\:a_n \;=\;-\frac{1}{2} + \frac{1+(\text{-}1)^n}{2}\cdot\frac{3}{4}$

$(3)\;8,\,4,\,8,\,4,\,8,\,4\;\hdots$

This sequence oscillates between 8 and 4.

The general term is: . $a_n \;=\;8 - \frac{1+(\text{-}1)^n}{2}\cdot4$

3. ## Re: Finding the nth term of the following sequence

Originally Posted by nisbahmumtaz
1) - 1/2 , 1/4 , - 1/2 , 1/4
It's pretty obvious isn't it? Assuming the indexing starts at 1, $a_n$ is -1/2 if n is odd, 1/4 if no is even. If you want a single "formula" note that $(-1)^n$ is -1 if n is odd and 1 if n is even. And that [itex]\frac{3}{8}- (-1)^n\frac{1}{8}[/tex] is $\frac{3}{8}+ \frac{1}{8}= \frac{1}{2}$ if n is odd, $\frac{3}{8}- \frac{1}{8}= \frac{1}{4}$ if n is even.

2) 11, 8, 13, 6, 15,...
Did you notice that 11- 3= 8, 8+ 5= 13, 13- 7= 6, 6+ 9= 15?

3) 8,4,8,4,8,4,...
Again this is just 8 if n is odd, 4 if n is even. That's a perfectly valid way of doing this. But if you want a "formula" again, it is $6- (-1)^n(2)$. I got that by noting that 6 is half way between 8 and 4, just as in (1) I saw that 3/8 is halfway between 1/4 and 1/2.

and could you guys share any tips on how to quickly find the nth term.
any help is much appreciated. thank you

4. ## Re: Finding the nth term of the following sequence

Hello again, nisbahmumtaz!

The second one is tricky . . .

$(2)\;11,\, 8,\, 13,\, 6,\, 15\,\hdots$

Look at the differences of consecutive terms.

. . $\begin{array}{|c|ccccccccc|} \hline _n & _1 && _2 && _3 && _4 && _5\\ \hline \text{Sequence} & 11 && 8 && 13 && 6 && 15 \\ \hline \text{Differences} && -3 && +5 && -7 && +9 & \\ \hline \end{array}$

Each term is: the previous term plus/minus a consecutive odd number.

We can write a recursive relation like this: . $a_{n+1} \;=\;a_n + (\text{-}1)^n(2n+1)$

Maybe someone can come up with a closed form?

5. ## Re: Finding the nth term of the following sequence

Originally Posted by Soroban
Maybe someone can come up with a closed form?
10-n(-1)^n

We should say of course that we can write down an nth term that gives all those terms and any number we please as the next term.

These are more like IQ test question than maths...?

6. ## Re: Finding the nth term of the following sequence

So as well as the simple $10- (-1)^n n$ we can also have $\frac{6n^4-70n^3+282n^2-455n+270}{3}$ which produces the sequence 11,8,13,6,15,116,433....