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?
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$\begingroup$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 $\displaystyle a_n=\color{red}{5}\dbinom{n-1}{0}+\color{red}{7}\dbinom{n-1}{1}+\color{red}{2}\dbinom{n-1}{2}=5+7(n-1)+2\frac{(n-1)(n-2)}{2}=\color{blue}{n^2+4n}$
$\endgroup$ $\begingroup$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 $(0,2), (1,6) , \text{ and } (2,10)$.
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