# How do you find the nth term formula 2, 5, 8, 10?

Aug 18, 2016

${a}_{n} = - \frac{1}{6} n \left(n + 1\right) \left(n - 7\right)$

#### Explanation:

Any finite number of terms does not determine a unique formula, unless you are given additional information about the sequence, e.g. that it is an arithmetic or geometric sequence.

Note that the given example is neither, since it has neither a common difference, nor a common ratio between successive terms.

We can derive an $n$th term formula by matching the terms with a polynomial, which can conveniently be done by examining sequences of differences...

Our original sequence is:

$\textcolor{b l u e}{2} , 5 , 8 , 10$

The sequence of differences of consecutive terms of that sequence is:

$\textcolor{b l u e}{3} , 3 , 2$

The sequence of differences of that sequence is:

$\textcolor{b l u e}{0} , - 1$

The sequence of differences of that sequence is:

$\textcolor{b l u e}{- 1}$

Having reached a constant sequence - albeit of just one term - we can then use the first term of each of these sequences as coefficients to give us a formula for the $n$th term:

a_n = color(blue)(2)/(0!) + color(blue)(3)/(1!)(n-1) + color(blue)(0)/(2!)(n-1)(n-2) + color(blue)(-1)/(3!)(n-1)(n-2)(n-3)

$= 2 + 3 n - 3 - \frac{1}{6} {n}^{3} + {n}^{2} - \frac{11}{6} n + 1$

$= \frac{1}{6} \left(- {n}^{3} + 6 {n}^{2} + 7 n\right)$

$= - \frac{1}{6} n \left({n}^{2} - 6 n - 7\right)$

$= - \frac{1}{6} n \left(n + 1\right) \left(n - 7\right)$

graph{ (y+1/6x(x+1)(x-7))((x-1)^2+(y-2)^2-0.02)((x-2)^2+(y-5)^2-0.02)((x-3)^2+(y-8)^2-0.02)((x-4)^2+(y-10)^2-0.02)=0 [-9, 14, -2, 12]}