# Technique for finding the nth term

I'm a trainee teacher and I need to pass a skills test in maths, but I'm having trouble in this particular area of algebra in finding the nth term of any sequence. I've been given a series of questions to answer one of them is:

Find the nth term of the sequence 3, 8, 15, 24.

I not only need to know the answer but more importantly the technique. This is GCSE level.

• 3 (+5) 8 (+7) 15 (+9) 24... – Angelo Rendina Dec 6 '14 at 13:35
• Isn't that squares minus 1 ? ($n^2 - 1$) – servabat Dec 6 '14 at 13:35
• Generally speaking, it's all but easy finding the closed formula for a recursive sequence, let alone if you only have few terms. First thing you may want to do is check if you can find a geometric/arithmetic progression. – Angelo Rendina Dec 6 '14 at 13:39

If you can't easily see the pattern, you can use OEIS to find him. Your specimen $3,8,15,24$ is small and in fact, lot of sequences could be used here.

As I can see it could be $(a_n)_{\mathbb{N}^+}$, such that $a_n = (n+1)^2-1 = n(n+2)$. $$\begin{split} 3 &=(1+1)^2-1&=1\cdot (1+2)\\ 8 &=(2+1)^2-1&=2\cdot (2+2)\\ 15 &=(3+1)^2-1&=3\cdot (3+2)\\ 24 &=(4+1)^2-1&=4\cdot (4+2) \end{split}$$ But it could be (in accordance with OEIS) something really else. Mostly the sequence apparent from the context.

When you are looking for a patter you should look at eg. the differences between successive numbers. Here $8-3 = 5, 15-8 = 7, 24-15=9$, so you can conclude $a_1 = 3 \wedge a_n = a_{n-1}+2n+1$ and if you want you can find that $a_n = n(n+2)$. Helpful should be looking for arithmetic/geometric/Fibonacci progression.

You can also look at difference between difference, as I made with other sequence and find the pattern, without any idea or context. Of course here is high chance of failure, but why not try it?

Neither the next term nor the general term of the whole sequence can be determined explicitly by just judging at a listing of a finite number of (the first few) terms.

I will use the given sample {3, 8, 15, 24, … } as an illustration of my point.

Suppose that the general term is "correctly guessed" as $T(n)$.

Try what sequence can be generated by $$T(n) + k(n – 1)(n – 2)(n – 3)(n – 4)$$; where $k$ is any non-zero arbitrary constant or even a simple function of n, like k = n.

It is therefore unfortunate and also unfair to require students (especial juniors) to guess the pattern by listing just the first few terms while there is no unique answer for it.