# Next Three Terms of a Quadratic Sequence

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### The Problem:

Find the next three terms in this sequence

14, 24, 36, 50, 66, ..., ..., ...,

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### Solution

As with any sequence we are given first let's look to see if it is not one we know. It's not one I know, so let's find the first difference between adjacent terms.

 $n$ 1 2 3 4 5 ${n}^{\mathrm{th}}$ term$\left({U}_{n}\right)$ 14 24 36 50 66 First difference $\left({U}_{n+1}-{U}_{n}\right)$ 10 12 14 16

This first difference is an arithmetic sequence with a common difference of 2, which enables us to work out the next three first differences:

 $n$ 1 2 3 4 5 6 7 8 ${n}^{\mathrm{th}}$ term$\left({U}_{n}\right)$ 14 24 36 50 66 ? ? ? First difference$\left({U}_{n+1}-{U}_{n}\right)$ 10 12 14 16 18 20 22

Adding these first differences in term to the previous term in the sequence we get:

 $n$ 1 2 3 4 5 6 7 8 ${n}^{\mathrm{th}}$ term$\left({U}_{n}\right)$ 14 24 36 50 66 84 104 126 First difference$\left({U}_{n+1}-{U}_{n}\right)$ 10 12 14 16 18 20 22

So the sequence is 14, 24, 36, 50, 66, 84, 104, 126,…

We can go further and find any term in the sequence. The first difference is an arithmetic sequence with a common difference of 2, implying that the original sequence is a quadratic sequence of the form ${U}_{n}=a{n}^{2}+bn+c$, where a is the common difference divided by 2, ∴ $a=1$.

 $n$ 1 2 3 4 5 6 7 8 ${n}^{\mathrm{th}}$ term$\left({U}_{n}\right)$ 14 24 36 50 66 84 104 126 First difference$\left({U}_{n+1}-{U}_{n}\right)$ 10 12 14 16 18 20 22 $a{n}^{2}$ 1 4 9 16 25 36 49 64 ${U}_{n}-a{n}^{2}$ 13 20 27 34 41 48 55 62

But ${U}_{n}-a{n}^{2}=bn+c$, so the last line is an arithmetic sequence with common difference $b=7$ and zeroth term $c=6$. Therefore, the ${n}^{\mathrm{th}}$ term of the sequence is given by ${U}_{n}={n}^{2}+7n+6$.

-oOo-