June 2022 Paper 1 Q12
12 The \(n\)th term of a sequence is \(\qquad \dfrac{3n^2}{n^2 + 2}\)
(a) One term in the sequence is \(\;\dfrac{32}{11}\)
Work out the value of \(n\). [2 marks]
(b) Write down the limiting value of the sequence as \(n \to \infty\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(33n^2 = 32(n^2 + 2)\) or \(\dfrac{64 - n^2}{11n^2 + 22} = 0\) | M1 | oe (both denominators should be cleared for the first method) |
| 8 | A1 | ignore \(-8\) in working as long as only 8 stated in answer |
Additional guidance
| May use T&I and will be 2 marks if they get the correct answer (0 marks without the answer) |
| Answer | Mark | Comments |
|---|---|---|
| 3 | B1 |
Additional guidance
| Condone \(\dfrac{3}{1}\) |