Higher June 2019 Paper 2R Q17
17 The table gives information about the first six terms of a sequence of numbers.
| Term number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Term of sequence | \(\dfrac{1 \times 2}{2}\) | \(\dfrac{2 \times 3}{2}\) | \(\dfrac{3 \times 4}{2}\) | \(\dfrac{4 \times 5}{2}\) | \(\dfrac{5 \times 6}{2}\) | \(\dfrac{6 \times 7}{2}\) |
Prove algebraically that the sum of any two consecutive terms of this sequence is always a square number.
(4)
| Scheme | Marks |
|---|---|
(Term \(n\) =) \(\dfrac{1}{2}n(n + 1)\) or (Term \(n + 1\) =) \(\dfrac{1}{2}(n + 1)(n + 2)\) | M1 |
| \(\dfrac{1}{2}n(n + 1) + \dfrac{1}{2}(n + 1)(n + 2)\) | M1 |
\(\dfrac{1}{2}(n + 1)(n + n + 2) = \dfrac{1}{2}(n + 1)(2n + 2)\) or \(\dfrac{1}{2}n^2 + \dfrac{1}{2}n + \dfrac{1}{2}n^2 + \dfrac{1}{2}n + n + 1 \to \underline{n^2 + 2n + 1}\) | M1 |
| Working required Answer: \((n + 1)^2\) shown | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Algebraic representation of one of the two consecutive terms in sequence
M1: Adding two consecutive terms
M1: Factorisation or multiplying out correctly to get to \(n^2 + 2n + 1\)
A1: Dep on M3