Arithmetic Progressions Formula

Question

I'm studying progressions in math class in Portugal, and I don't know the words/translation in english for certain things so I'll try to explain.

I have this arithmetic progression: 2, 8, 18, 32 ...

I want to know the 5th term. Is there a formula to know the general term(the formula to know any of the terms) without the reason(r)? I read online that: Un = U1 + (n-1) * r

r is the difference between Un+1 and Un. (Un+1 - Un)

In this progression though, the r changes, so I'm guessing I can't use that formula.

Answer 1

A true arithmetic progression does follow the rule $U_n=U_1+(n-1)r$, so this is not an arithmetic progression. There is no general rule for finding the general term of a sequence. If the sequence is the list of values of a polynomial, taking the difference of successive terms will result in a constant eventually. In your case:$$2\quad 8 \quad 18 \quad 32 \\ \ \ 6 \quad 10 \quad 14 \\ \quad 4 \quad 4$$ We now assume that the $4$'s continue. The fact that it takes two sets of differences says the polynomial is of second degree. The leading term is the final difference divided by the factorial of the degree, so here $\frac 4{2!}=2$ and the leading term of he polynomial is $2n^2$ You can subtract that off from the values to get a polynomial of one degree lower. In this case that is all there is.

Answer 2

Let's try to acquire general term

By writing sequence recursively we have $$\begin{align}a_1&=2\\ a_2&=a_1+(6+4(1-1))=8\\ a_3&=a_2+(6+4(3-2))=18\\ a_4&=a_3+(6+4(4-2))=32\\ \vdots\\ a_n&=a_{n-1}+(6+4(n-2))\\ a_n&=2+\sum_{k=1}^{n-1}{(6+4(k-2))} \end{align}$$ (By summing all steps)

$$a_n=2+2(n-1)+4\frac{n(n-1)}{2}$$

$$a_n=\frac{4+4(n-1)+4n(n-1)}{2}$$

$$a_n=\frac{4+4n-4+4n^2-4n}{2}=\frac{4n^2}{2}=2n^2$$

General term of this sequence is

$$2n^2$$