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First of all, I am a 12th grader so I don't know how to write research notes. So please forgive me if my writing is not so impressive!

I don't know what to do to tell the world about whatever I found. So, may be it's stupid, but I am posting my formula here in Stack Exchange.

Recently I was trying to figure out how to find the nth term in a series, or sequence, where the differences ($d$) between the elements of the series are not constant, but if $d$ for each $A_n - A_{(n-1)}$ is written in a sequence, it gives an AP.

For example, the sequence:

$A = 7, 14, 28, 49, ... n$

is not in AP

But taking the differences for all $A_n - A_{(n - 1)}$, where $n > 1$, we get an AP:

$$A' = 7, 14, 21... $$

The above AP has a common difference of 7.

In order to find $A_n$, We can use the below formula:

$$A_n = \left(\frac {n - 1}{2} \{2a' + (n - 2) d' \}\right) + a_1$$

Where,

$a_1$ is the first element in the sequence

$$a' = a_2 - a_1$$

$$d' = a_3 - (a' + a_2)$$

I can't describe here how I get to that point.

Here is the application of the formula in the following sequence:

$$A = 7, 14, 28, 49, ... n$$

In sequence, let $n = 3$ and we have to find $A_n$

$$a_1 = 7$$

$$a' = 14 - 7 = 7$$

$$d' = 28 - (14 + 7) = 7$$

By formula:

$$A_n = \frac {n - 1}{2} \{2a' + (n - 2) d' \} + a_1$$

$$A_3 = \frac {3 - 1}{2} \{2 \times 7 + (3 - 2) 7 \} + 7$$

$$A_3 = \frac {2}{2} \{14 + 7 \} + 7 = 28$$

Notes:

1) I am still testing it.

2) There are other methods exist to find $A_n$ in these type of sequences, may be they are easier, I don't know.

3) I am not sure if I am the first person to derive this formula, if I am not, then I apologize.

Let me know, if someone has found something wrong with this.

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Your method, while interesting, is not new: this is simply an example of a recurrence relation. An easier way to solve this is to first note that two subsequent subtractions yields a constant (see link in comments above).

Now we set up equations in quadratic form:
$$a(1)^2+b(1)+c = 7 \implies a+b+c = 7\\ a(2)^2+b(2)+c = 14 \implies 4a+2b+c=14\\ a(3)^2+b(3)+c=28 \implies 9a+3b+c = 28$$
Solving this, we get the quadratic $\frac{7}{2}x^2 -\frac{7}{2}x + 7$ which includes all of your points (specifically integers greater than or equal to $1$)

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    $\begingroup$ That's easier. Huff! I am dumb. Thanks :) $\endgroup$ – Devashish Jun 30 '16 at 21:07
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    $\begingroup$ You are not dumb. You are doing something new to you, and new work is always awkward. The important thing is that you did this by yourself, and that is good. $\endgroup$ – marty cohen Jun 30 '16 at 22:55
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    $\begingroup$ @Devashish I second marty on this one. While you may not have taken the most direct route, your work definitely shows dedication, time, and a sharp mind. Keep it up... these are qualities fundamental to good mathematics! $\endgroup$ – Brevan Ellefsen Jul 1 '16 at 2:03
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You can take this farther: Let $A^{(1)}_n=A^{(0)}_{n+1}-A^{(0)}_n$ and $A^{(2)}=A^{(1)}_{n+1}-A^{(1)}_n$ and $A^{(3)}_n=A^{(2)}_{n+1}-A^{(2)}_n$ and so on. Then $A^{(j)}_n$ is a non-zero constant if and only if $A^{(0)}_n$ is a polynomial function of $n,$ with degree $j.$ You can prove this by induction on $j.$

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