Is there any formula to find the nth element in a sequence where common difference (d) is varying with a constant rate?

Question

To explain my question, here is an example.

Below is an AP:

2, 6, 10, 14....n

Calculating the nth term in this sequence is easy because we have a formula. The common difference (d = 4) in AP is constant and that's why the formula is applicable, I think.

But what about this sequence:

5, 12, 21, 32....n

Here, the difference between two consecutive elements is not constant, but it too has a pattern which all of you may have guessed. Taking the differences between its consecutive elements and formimg a sequence results in an AP. For the above example, the AP looks like this:

5, 7, 9, 11.....n

So given a sequence with "uniformly varying common difference" , is there any formula to calculate the nth term of this sequence?

Answer 1

Here's how you can find this with a difference table:

$\color{red}{5}\;\;\; 12\;\;\;21 \;\;\;32\;\cdots$

$\;\;\color{red}{7}\;\;\;\;9\;\;\;\;11\;\cdots$

$\;\;\;\;\color{red}{2}\;\;\;\;2\;\;\;\cdots$

$\hspace{.33 in}0$

So $\displaystyle a_n=\color{red}{5}\dbinom{n-1}{0}+\color{red}{7}\dbinom{n-1}{1}+\color{red}{2}\dbinom{n-1}{2}=5+7(n-1)+2\frac{(n-1)(n-2)}{2}=\color{blue}{n^2+4n}$

Answer 2

Well the explicit form aka the nth term is a quadratic... $A_n=an^2+bn+c$

Use the points from your sequence to find $a,b, \text{ and } c$

For example you can use the points $(0,2), (1,6) , \text{ and } (2,10)$.