Question

Find a formula for the general term a_n of the sequence {5,1,5,1,5,1...}

Answer 1

look like
\[a_n= \left\{\begin{array}{rcc}
5 & \text{if n is odd} & \\
1& \text{if n is even } &
\end{array}
\right.
\]

Answer 2

something like : 3+2*(-1)^n

Answer 3

How about \(5^{n} \pmod {12} \) ?

Answer 4

3+2*(-1)^n is correct!!! I didnt even think of 3 and 2, smh

Answer 5

@foolformath wow, just wow... how do you do it, I don't know lol

Answer 6

Yeah, what does that mean?

Answer 7

Even mod 24 works ;)

Answer 8

What does "mod" mean??

Answer 9

for some reason that escape my puny brain, 5n mod 12 alternate between 1 and 5. probably because 5 mod 12 = 5 and 25 mod 12 = 1, if you multiply 25 mod 12 by 5, you get 5 again? I dunno...

Answer 10

Yes that's right.

Answer 11

mod is basically the remainder operation here.

Answer 12

that's the remainder of your division. for exemple, 5 mod 4 = 1 because you'll be left with 1 after your division. 25 mod 12 is 1 as well because,once you've removed 24 from 25, you're left with 1

Answer 13

0___0
never really understood that remainder stuff. But okay. If i go on to Calc 3, i'll definitely use this.

Answer 14

oh, it's not really calc stuff, it's just something that's generally used more in programming than in math (at least, the maths I did ;-) ). It's not a much used arithmetic operation.

Answer 15

I had it in my computer class, but i never understood how he always got "remainder 1".

Answer 16

Another beautiful pattern,

\[\frac{1}{66} = 0.015151515151515151515151515151515151515\cdots \]

Answer 17

thats recursive, right?

Answer 18

I will assume that you mean 'recurring'.

Answer 19

or 1 + 4 * n mod 2. For every even number, n mod 2 = 0, so an = 1. Every odd number, n mod 2 = 1, so an = 5

Answer 20

@FoolForMath Yes, i think thats what I mean. If I stay with math for a long time, my vocabulary/grammar plummets