Higher June 2023 Paper 2 Q15
15 Here are the first four terms of a quadratic sequence.
3 9 17 27
Find an expression, in terms of \(n\), for the \(n\)th term of this sequence. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(n^2 + 3n - 1\) | M1 | for a correct start to a method to find the \(n\)th term, eg constant 2nd differences and \(n^2\) OR states \(2a = 2\) or \(3a + b = 6\) |
| M1 | for working with \(n^2\), eg \(n^2\) and sequence 2, 5, 8, … OR states \(2a = 2\) and \(3a + b = 6\) | |
| A1 | for \(n^2 + 3n - 1\) |
Additional guidance
Need to see constant second difference found and \(n^2\)
Condone use of different variable throughout
\(a = 1\) or \(b = 3\) implies M1
\(n^2 + 3n\) implies M2
\(a = 1\) and \(b = 3\) implies M2