Sequences

Algebra and Sequences

When the sequence is not simply the squares shifted, the second differences still give the start of the formula
Halve the second difference: that number multiplies n²
Write out that multiple of the square numbers, then subtract it term by term from the original sequence
What is left is always a linear sequence, so find its nth term by the method in section 3
Adding the two parts together gives the complete nth term

Worked example

A quadratic with three parts

Find the nth term of 3, 6, 13, 24, 39.

Solution:

First differences: 3, 7, 11, 15
Second differences: 4, 4, 4, so the sequence is quadratic
Half of 4 is 2, so the formula contains 2n²
Write out 2n²: 2, 8, 18, 32, 50
Subtract term by term: 3 − 2 = 1, 6 − 8 = −2, 13 − 18 = −5, 24 − 32 = −8
Those leftovers are 1, −2, −5, −8, a linear sequence with step −3 and previous term 4, giving 4 − 3n
Adding the two parts: the nth term is 2n² − 3n + 4
Check with n = 5: 50 − 15 + 4 = 39, which matches

The same routine handles a cubic, except that the constant third difference is divided by 6 to give the multiple of n³
- For 2, 15, 52, 125, 246 the third differences are 12, so the formula contains 2n³, and subtracting 2n³ leaves 0, −1, −2, −3, whose nth term is 1 − n, making the answer 2n³ − n + 1