# Thread: Determining the general term

1. ## Determining the general term

The general term is given by 3r+1+(-1)^r+1

So by substituting r=0,1,2,3.... I get a sequence like this: 0, 5, 6, 11, 12, 17, 18.....

It seems to form some pattern. So i wonder, can i deduce the general term with only the sequence? How?

Please kindly elaborate more if possible because i really keen to learn how.

2. ## Re: Determining the general term

Originally Posted by MichaelLight
The general term is given by 3r+1+(-1)^r+1

So by substituting r=0,1,2,3.... I get a sequence like this: 0, 5, 6, 11, 12, 17, 18.....

It seems to form some pattern. So i wonder, can i deduce the general term with only the sequence? How?

Please kindly elaborate more if possible because i really keen to learn how.
The sequence can be described as follows:
Consider the rth term (r starts from 0).
If r is even, $a_r = 3r$.
If r is odd, $a_r = 3r+2$.
So, we need to subtract 1 from 3r+1 if r is even and add 1 to 3r+1 if r is odd. An easy way to do this would be to add $(-1)^{r+1}$ to 3r+1.
Thus, we find that the general term is given by $a_r=3r+1+(-1)^{r+1}$.

3. ## Re: Determining the general term

Hello, MichaelLight!

The general term is given by: . $a_n \:=\: 3n+1+(\text{-}1)^{n+1}$

So by substituting $r\:=\:0,1,2,3\,\hdots$ I get a sequence like this: $0, 5, 6, 11, 12, 17, 18 \hdots$

It seems to form some pattern.
So i wonder, can i deduce the general term from the sequence?

When given an increasing sequence of terms, I usually take the difference
. . of consecutive terms. Then the differences of the differences, and so on.
We hope to find that the $n^{th}$ differences are constant.
. . This indicates that the generating function is an $n^{th}$ degree polynomial.

So we have:

. . $\begin{array}{c|cccccccccccccc} \text{Sequence} & 0 && 5 && 6 && 1 && 12 && 17 && 18 \\ \text{1st diff.} && 5 && 1 && 5 && 1 && 5 && 1 \\ \text{2nd diff.} &&& \text{-}4 && 4 && \text{-}4 && 4 && \text{-}4 \end{array}$

But we see that the 1st difference has alternating terms.
This indicates that the sequence is composed of two subsequences,
. . one for "even" terms, one for "odd" terms.

We have:

. . $\begin{array}{c||c|c|c|c|c|c|c|c|}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{even} & 0 && 6 && 12 && 18 \\ \hline \text{odd} && 5 && 11 && 17 & \\ \hline \end{array}$

We see that, for even $n:\;a_n \:=\:3n$
. . . . . . . . . . for odd $n:\;a_n \:=\:3n+2$

And from here, we can consult Alex's excellent explanation.