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 :)

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- Jul 14th 2012, 06:53 PMnisbahmumtazFinding 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 :) - Jul 14th 2012, 07:34 PMSorobanRe: 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.

Quote:

$\displaystyle (1)\;\text{-}\tfrac{1}{2},\;\tfrac{1}{4},\;\text{-}\tfrac{1}{2},\;\tfrac{1}{4},\;\hdots$

$\displaystyle \text{When }n\text{ is odd, }b(n) = 0.$

$\displaystyle \text{When }n\text{ is even, }b(n) = 1.$

. . I call it the "blinker" function.

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

Quote:

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

This sequence oscillates between 8 and 4.

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

- Jul 14th 2012, 07:52 PMHallsofIvyRe: Finding the nth term of the following sequence
It's pretty obvious isn't it? Assuming the indexing starts at 1, $\displaystyle a_n$ is -1/2 if n is odd, 1/4 if no is even. If you want a single "formula" note that $\displaystyle (-1)^n$ 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 $\displaystyle \frac{3}{8}+ \frac{1}{8}= \frac{1}{2}$ if n is odd, $\displaystyle \frac{3}{8}- \frac{1}{8}= \frac{1}{4}$ if n is even.

Quote:

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

Quote:

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

Quote:

and could you guys share any tips on how to quickly find the nth term.

any help is much appreciated. thank you :)

- Jul 15th 2012, 09:56 AMSorobanRe: Finding the nth term of the following sequence
Hello again, nisbahmumtaz!

The second one is tricky . . .

Quote:

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

Look at the differences of consecutive terms.

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

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

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

Maybe someone can come up with a closed form?

- Jul 15th 2012, 10:42 AMa tutorRe: Finding the nth term of the following sequence
- Jul 19th 2012, 01:08 AMa tutorRe: Finding the nth term of the following sequence
So as well as the simple $\displaystyle 10- (-1)^n n$ we can also have $\displaystyle \frac{6n^4-70n^3+282n^2-455n+270}{3}$ which produces the sequence 11,8,13,6,15,116,433....