Higher November 2019 Paper 1 Q13
13 The \(n\)th term of a sequence is \(\quad \dfrac{n(n - 4)}{\sqrt{n + 3}}\)
Work out the sum of the 1st and 6th terms. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{1(1 - 4)}{\sqrt{1 + 3}}\) or \(\dfrac{-3}{\sqrt{4}}\) or \(\dfrac{6(6 - 4)}{\sqrt{6 + 3}}\) or \(\dfrac{6 \times 2}{\sqrt{9}}\) or \(\dfrac{12}{3}\) or \(\dfrac{4}{1}\) | M1 | oe eg \(\dfrac{1^2 - 1 \times 4}{\sqrt{1 + 3}}\) eg \(\dfrac{6^2 - 6 \times 4}{\sqrt{6 + 3}}\) |
| \(\dfrac{-3}{2}\) or \(-1\dfrac{1}{2}\) or \(-1.5\) or 4 | M1dep | |
| \(2\dfrac{1}{2}\) or \(\dfrac{5}{2}\) or 2.5 | A1 | oe mixed number, fraction or decimal |
Additional guidance
| \(\dfrac{n^2 - 4n}{\sqrt{n + 3}}\) with no correct substitution | M0M0A0 |