# How do you find the nth term of the sequence 2,5,10,17,26,37,...?

Feb 13, 2017

the nth term is ${n}^{2} + 1$

#### Explanation:

The sequence above is quadratic, and we use the expression $a {n}^{2} + b n + c$ where 'a' represents $\frac{1}{2}$ the second difference and 'c' is the 0th term. We then substitute numbers into the equation to find the value of 'b'.

For example, the second difference in your sequence would be 2 because the 1 first differences are 3, 5, 7, 9 and 11. Thus 'a' = 1 because $\frac{2}{2} = 1$. Then we work backwards to find 'c'. The first difference for the first two terms is 3. $3 - 2 = 1$. Then the first term - 1 = 1. After this we just substitute numbers into the equation to find the value of 'b'.

$1 {n}^{2} + 1 b + 1 = 2$, when n = 1
$2 + b = 2$
$b = 0$
so the nth term is ${n}^{2} + 1$

Feb 14, 2017

${n}^{2} + 1$

#### Explanation:

This is the GCSE approach:

Take the difference between consecutive terms, and then take the difference between those terms, as follows:

$\left.\begin{matrix}\text{Seq: " & 2 & \null & 5 & \null & 10 & \null & 17 & \null & 26 & \null & 37 & \null \\ "1st Difference:" & \null & 3 & \null & 5 & \null & 7 & \null & 9 & \null & 11 & \null & \null \\ "2nd difference:} & \null & \null & 2 & \null & 2 & \null & 2 & \null & 2 & \null & \null & \null\end{matrix}\right.$

So as the 2nd difference is constant, we can infer that the terms form a quadratic sequence, of the form:

$\left\{{u}_{n}\right\} = a {n}^{2} + b x + c$

and the quadratic coefficient is twice the 2nd difference, ie:

$2 a = 2 \implies a = 1$

We now form a table where we take the original sequence and subtract the quadratic term to give a residue:

$\left.\begin{matrix}n : & 1 & 2 & 3 & 4 & 5 & 6 \\ \text{Seq ("n"): " & 2 & 5 & 10 & 17 & 26 & 37 \\ an^2=n^2: & 1 & 4 & 9 & 16 & 25 & 36 \\ "Residue:} & 1 & 1 & 1 & 1 & 1 & 1\end{matrix}\right.$

So we can now also conclude that the linear part of the sequence ($b x + c$) form the terms $\left\{1 , 1 , 1 , 1 , \ldots\right\}$, so we can conclude that $b = 0$ and $c = 1$ and hence the general term of the sequence is:

$\setminus \setminus \setminus \setminus \setminus {u}_{n} = {n}^{2} + 0 n + 1$
$\therefore {u}_{n} = {n}^{2} + 1$

Which we can verify as follows:

$\left.\begin{matrix}n : & 1 & 2 & 3 & 4 & 5 & 6 \\ {n}^{2} : & 1 & 4 & 9 & 16 & 25 & 36 \\ {n}^{2} + 1 & 2 & 5 & 10 & 17 & 26 & 37\end{matrix}\right.$