Higher June 2019 Paper 3 Q16
16 Here are the first six terms of a quadratic sequence.
\(-1 \qquad 5 \qquad 15 \qquad 29 \qquad 47 \qquad 69\)
Find an expression, in terms of \(n\), for the \(n\)th term of this sequence. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2n^2 - 3\) | M1 | begins to work with 2nd differences \(6 \quad 10 \quad 14 \quad 18 \quad 22\) \(\ \ 4 \quad\ \ 4 \quad\ \ 4 \quad\ \ 4\) |
| M1 | identifies \(2n^2\) as part of the expression eg gives the sequence 2, 8, 18, 32, ... or gives a quadratic expression which includes the term \(2n^2\) | |
| A1 | oe |
Additional guidance
A quadratic expression of the form \(2n^2 + bn + c\) can be awarded the first 2 marks