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
Hello, nisbahmumtaz!
(1) and (3) are oscillating sequences.
There are several ways to express the general term.
I'll show you one of them.
Consider the expression: .$\displaystyle b(n) \:=\:\frac{1 + (\text{-}1)^n}{2}$$\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.
$\displaystyle \text{Therefore, the }n^{th}\text{ term is: }\:a_n \;=\;-\frac{1}{2} + \frac{1+(\text{-}1)^n}{2}\cdot\frac{3}{4}$
$\displaystyle (3)\;8,\,4,\,8,\,4,\,8,\,4\;\hdots$
This sequence oscillates between 8 and 4.
The general term is: .$\displaystyle a_n \;=\;8 - \frac{1+(\text{-}1)^n}{2}\cdot4$
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.
Did you notice that 11- 3= 8, 8+ 5= 13, 13- 7= 6, 6+ 9= 15?2) 11, 8, 13, 6, 15,...
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 $\displaystyle 6- (-1)^n(2)$. 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 1/2.3) 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
Hello again, nisbahmumtaz!
The second one is tricky . . .
$\displaystyle (2)\;11,\, 8,\, 13,\, 6,\, 15\,\hdots$
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?