Question

is it possible to find the nth term for the series
1,5,7,11,13,17,19,....

Answer 1

no, i came across this series when i had to find the nth harmonics in motor current excluding the even harmonics and multiples of 3 harmonics

Answer 2

Well its pretty vague, you can't determine the nth term of a sequence of terms just by giving a few of them

Answer 3

You can guess tho lol

Answer 4

well , 6n+1 gives the series 1,7,13,....
and 6n-1 gives 5,11,17,....
i want them both in a single expression

Answer 5

Then you should have just asked for that in the first place.

Answer 6

well i m sorry... i was trying to explain that i didn't them from working on prime numbers

Answer 7

You want $$\{ 6n-1 , n\in \mathbb{N}\}\bigcup \{ 6n+1 , n\in \mathbb{N}\}$$

Answer 8

you got it....
but how do i put it as a single expression?

Answer 9

I just did

Answer 10

a single expression which can be put as a statement in a program.... which will give the series 1,5,7,11,13,.... for each successive value of n (where n is a whole number)

Answer 11

k one sec

Answer 12

it seems your asking for a solution to 6n+1=6n-1

Answer 13

take your time buddy

Answer 14

@abhinayreddy here it is i got it

Answer 15

my bad one sec

Answer 16

that worked, you just had to start at n = -1, very cool....

Answer 17

@abhinayreddy $$\frac{1}{2}(6n-3+(-1)^n)$$

Answer 18

@jack119how did you arrive at that expression.....?

Answer 19

please explain me.... and thanks a lot man