A2 June 2020 Paper 2 Q10
10 The sequence \(u_1, u_2, u_3, \ldots\) is defined by
\[u_1 = 0 \qquad u_{n+1} = \frac{5}{6 - u_n}\]Prove by induction that, for all integers \(n \geqslant 1\),
\[u_n = \frac{5^n - 5}{5^n - 1}\][6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Shows that \(u_n = \dfrac{5^n - 5}{5^n - 1}\) is true for \(n = 1\) | B1 | 1.1b |
| States the assumption that \(u_n = \dfrac{5^n - 5}{5^n - 1}\) is true for \(n = k\) | M1 | 2.4 |
| Uses the recurrence relation and the assumption to express \(u_{k+1}\) in terms of \(k\) | M1 | 3.1a |
| Expresses \(u_{k+1}\) as a single fraction. | M1 | 1.1a |
| Completes rigorous working to deduce that \(u_{k+1} = \dfrac{5^{k+1} - 5}{5^{k+1} - 1}\) | R1 | 2.2a |
| Concludes a reasoned argument by stating that the formula for \(u_n\) is true for \(n = 1\); that if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction \(u_n = \dfrac{5^n - 5}{5^n - 1}\) for \(n \geqslant 1\) | R1 | 2.1 |
| (6 marks) |
Typical solution
Let \(n = 1\); then the formula gives
\[u_1 = \frac{5^1 - 5}{5^1 - 1} = 0\]so the result is true for \(n = 1\)
Assume the result is true for \(n = k\):
Then \(u_{k+1} = \dfrac{5}{6 - \left(\dfrac{5^k - 5}{5^k - 1}\right)}\)
\[\begin{aligned} 6 - \left(\frac{5^k - 5}{5^k - 1}\right) &= \frac{6(5^k - 1) - (5^k - 5)}{5^k - 1} \\ &= \frac{6 \times 5^k - 5^k - 6 + 5}{5^k - 1} \\ &= \frac{5 \times 5^k - 1}{5^k - 1} = \frac{5^{k+1} - 1}{5^k - 1} \end{aligned}\]\[\therefore u_{k+1} = 5 \times \frac{5^k - 1}{5^{k+1} - 1} = \frac{5^{k+1} - 5}{5^{k+1} - 1}\]and the result also holds for \(n = k + 1\)
The formula for \(u_n\) is true for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction \(u_n = \dfrac{5^n - 5}{5^n - 1}\) for \(n \geqslant 1\)