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.
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$\begingroup$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?
$\endgroup$ $\begingroup$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.
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