How to find the nth term of a quadratic sequence

A sequence is quadratic when the gaps between terms change, but change by the same amount each time. That constant second difference is the key to the whole method.

See the second difference first

Fill in the hidden terms, then check the difference ladder underneath. The gaps grow, but the gaps between the gaps do not.

The method, in four steps

1. Work out the first differences, then the second differences.
2. Halve the second difference to get $a$, the number in front of $n^2$.
3. Subtract $a{n}^2$ from each term.
4. What is left is linear. Find its nth term and add it on.

Example 1

Find the nth term of $3,\ 8,\ 15,\ 24,\ 35$.

Step 1. First differences: $5, 7, 9, 11$.Seconddifference:$2$.

Step 2. Half of 2 is 1, so the sequence starts with $n^2$.

Step 3. Subtract $n^2$ (which is $1, 4, 9, 16, 25$) from each term:

Step 4. $2, 4, 6, 8$islinearwithnthterm$2n$. So:

Check: $n=5$ gives $25 + 10 = 35$.

Example 2

Find the nth term of $5,\ 11,\ 21,\ 35,\ 53$.

First differences: $6, 10, 14, 18$.Seconddifference:$4$,so$a = 2$andthesequencestartswith$2n^2$.

Subtract $2n^2$ (which is $2, 8, 18, 32, 50$):everyresultis$3$. The leftover is constant, so:

Check: $n=5$ gives $50 + 3 = 53$.

Mistakes that cost marks

1. Using the second difference as $a$ instead of half of it.
2. Subtracting $n^2$ instead of $a{n}^2$ when $a$ is not 1.
3. Not checking a term at the end. One substitution catches most slips.