Question

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

Answer 1

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

$\begin{cases}a_1=14\\a_2=37\\a_3=72\\a_4=119\\a_5=178\end{cases}$

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

The usefulness is this: if the sequence were arithmetic, then the forward differences would be constant. For example, if the sequence were $\{1,2,3,\ldots\}$, we would see the differences to be $\{1,1,\ldots\}$, 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 $c_i=b_{i+1}-b_i$.

$\implies\begin{cases}c_1=35-23=12\\c_2=47-35=12\\c_3=59-47=12\end{cases}$

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

$b_4=b_3+12=b_2+24=b_1+36$

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

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

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

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

There are lots of ways to find the explicit formula for $a_n$ 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

$a_n=an^2+bn+c$

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

$n=1\implies14=a+b+c$
$n=2\implies37=4a+2b+c$
$n=3\implies72=9a+3b+c$

$\implies a=6,b=5,c=3$

So the $n$-th term is determined by

$a_n=6n^2+5n+3$

for all $n\ge1$.