Finding general term of sequence

Question

I need to find the general terms of the sequence, $$\{1.25, 2, 3.25, 5, 7.25, \cdots\}$$ and $$\{0, 6, 24, 60, 120, 210, \cdots\}$$

I can't think of any such pattern that would produce these.

Answer 1

For first one, see the differences, the differences are $0.75,1.25,1.75,2.25,\cdots$ and further their differences are of $0.5$. So I hope, you can write the general term from here.

As @Ethan points out in the comments, mathematicians prefer fractions over decimals, one should convert the sequence in the form $$\dfrac{5}{4},\dfrac{8}{4},\dfrac{13}{4},\dfrac{20}{4},\cdots$$ So it is easy to spot that numerators increase by $3,5,7,9,\cdots$ and denominator is constant so it is easy to write the general term from this way also (in fact more easier to see after converting in fractions).

For second one, it is easy to observe that the terms are $1^3-1,2^3-2,3^3-3,4^3-4,\cdots$.

Answer 2

With a problem like this there are some standard things to try. One is forming the sequence of differences, as in Light Yagami’s answer. If they’re constant, the original sequence is arithmetic. If not, you can look at the differences of the difference sequence, and if they’re constant, the original sequence is described by a quadratic. And so on: if the $n$-th difference sequence is constant, the original sequence is described by a polynomial of degree $n$.

Instead of differences, you can look at the ratios of consecutive terms; if those seem to be nearly constant, the sequence is likely to be some modification of a geometric sequence, e.g., something like $a_n=3\cdot2^n-4n+1$.

Another possibility is seeing whether multiplying the sequence by some constant, or adding some constant to each term, results in a familiar sequence. The more sequences you’re familiar with, the more useful this can be. For instance, for the second sequence one might also notice that each term is a multiple of $6$. Dividing out that common factor leaves a sequence $\langle 0,1,4,10,20,35,\ldots\rangle$, and if one has done enough with binomial coefficients and Pascal’s triangle, one might notice that this is the sequence

$$\left\langle\binom23,\binom33,\binom43,\binom53,\binom63,\binom73,\ldots\right\rangle\,.$$

Thus, the sequence $\langle 0,6,24,60,120,210,\ldots\rangle$ is

$$\left\langle 6\binom{n}3:n\ge 2\right\rangle=\langle n(n-1)(n-2):n\ge 2\rangle$$

which you can write as

$$\left\langle 6\binom{n+1}3:n\ge 1\right\rangle=\langle n(n+1)(n-1):n\ge 1\rangle$$

if you prefer to index the terms starting at $n=1$, or as

$$\left\langle 6\binom{n+2}3:n\ge 0\right\rangle=\langle n(n+1)(n+2):n\ge 0\rangle$$

if you prefer $0$-based indexing. With $1$-based indexing you get the nice closed form

$$a_n=n(n+1)(n-1)=n^3-n\,,$$

as Light Yagami found by a different route.

Similarly, after converting the first sequence to fractions with a common denominator, one might notice that the sequence $\langle 5,8,13,20,29,\ldots\rangle$ of numerators is $4$ more than a sequence of consecutive squares:

$$\left\langle1^2+4,2^2+4,3^2+4,4^2+4,5^2+4,\ldots\right\rangle\,.$$

The $1$-based general form would therefore appear to be

$$a_n=\frac{n^2+4}4=\left(\frac{n}2\right)^2+1\,.$$