GCSE(F), GCSE(H),

A quadratic sequence is given by `U_n=an^2+bn+c`, where `a, b text( and ) c` are constants, `n` is the term and `U_n` is the value of the term. Note that there is no higher power than `n^2` in a quadratic sequence.

The second difference of a quadratic sequence is a constant: divide the second difference by 2 to obtain the coefficient of the `x^2` term.

To work out the quadratic sequence, determine the `an^2` value of each term; then subtract that from the values of the original sequence to obtain a linear sequence. Solve the linear sequence based on `bn + c`.

1. What is the *n*^{th} term of the quadratic sequence given by 3, 12, 27, 48, 75, ...?

Answer: `U_n=3n^2+4n+5`

Work out the second differences for the first five terms:

Term | 1 | 2 | 3 | 4 | 5 | ... | |||||
---|---|---|---|---|---|---|---|---|---|---|---|

Value | 12 | 25 | 44 | 69 | 100 | ... | |||||

1^{st} Difference | 13 | 19 | 25 | 31 | ... | ||||||

2^{nd} Difference | 6 | 6 | 6 | ... |

The second difference is 6; the multiple for `n^2` is 6 รท 2 = 3.

Subtract the value of `3n^2` from the original sequence:

Term | 1 | 2 | 3 | 4 | 5 | ... | |||||
---|---|---|---|---|---|---|---|---|---|---|---|

Original | 12 | 25 | 44 | 69 | 100 | ... | |||||

3n^{2} | 3 | 12 | 27 | 48 | 75 | ... | |||||

Original - 3n^{2} | 9 | 13 | 17 | 21 | 25 | ... | |||||

Difference | 4 | 4 | 4 | 4 | ... |

The difference is 4, to give 4*n* as that part of the sequence.

Work out the value of the zero term: 9 - 4 = 5. Assemble the parts: `U_n = 3n^2 + 4n + 5`

2. What is the second term of the sequence `U_n=n^2-n+1`?

Answer: 91

Substitute for n with 10 in the sequence: `10^2 - 10 + 1` = 91`.

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