Find the nth term of the sequence:14, 37, 72, 119, 178

Check the forward differences: if [tex]a_i[/tex] is the [tex]i[/tex]-th term, then the first forward difference of [tex]a_i[/tex] is [tex]b_i=a_{i+1}-a_i[/tex].

[tex]\begin{cases}a_1=14\\a_2=37\\a_3=72\\a_4=119\\a_5=178\end{cases}[/tex]

[tex]\implies\begin{cases}b_1=27-14=23\\b_2=72-37=35\\b_3=119-72=47\\b_4=178-119=59\end{cases}[/tex]

The usefulness is this: if the sequence were arithmetic, then the forward differences would be constant. For example, if the sequence were [tex]\{1,2,3,\ldots\}[/tex], we would see the differences to be [tex]\{1,1,\ldots\}[/tex], and so the sequence would be growing linearly, or by an added constant.

In our case, we can compute the forward differences again (the second-order differences) until we find such a pattern. This time, we denote it by [tex]c_i=b_{i+1}-b_i[/tex].

[tex]\implies\begin{cases}c_1=35-23=12\\c_2=47-35=12\\c_3=59-47=12\end{cases}[/tex]

So now we know that the sequence [tex]b_n[/tex] is arithmetic with a common difference of 12 between its terms.

[tex]b_4=b_3+12=b_2+24=b_1+36[/tex]

Now consider [tex]a_3=72[/tex]; we can write this term of the sequence in terms of the previous ones. We have

[tex]35-23=12\implies 35=23+12[/tex]
[tex]\implies72-37=23+12[/tex]
[tex]\implies\underbrace{72}_{a_3}=\underbrace{37}_{a_2}+\underbrace{23}_{a_2-a_1=37-14}+12[/tex]
[tex]\implies a_3=2a_2-a_1+12[/tex]

We would see a similar recursive pattern if we looked at the other terms [tex]a_4=119[/tex] and [tex]a_5=178[/tex]. We then establish that the sequence is given by the recursive formula

[tex]\begin{cases}a_1=23\\a_2=35\\a_{n+2}=2a_{n+1}-a_n+12&\text{for }n>2\end{cases}[/tex]

There are lots of ways to find the explicit formula for [tex]a_n[/tex] from this point, but the simplest is to realize that the sequence must be quadratic (this is because the second-order differences are constant) so we can assume that

[tex]a_n=an^2+bn+c[/tex]

for some constants [tex]a,b,c[/tex]. We have to solve for three unknowns, so we need three known values:

[tex]n=1\implies14=a+b+c[/tex]
[tex]n=2\implies37=4a+2b+c[/tex]
[tex]n=3\implies72=9a+3b+c[/tex]

[tex]\implies a=6,b=5,c=3[/tex]

So the [tex]n[/tex]-th term is determined by

[tex]a_n=6n^2+5n+3[/tex]

for all [tex]n\ge1[/tex].