FP1 January 2009 Q6
6. A series of positive integers \(u_1, u_2, u_3, \ldots\) is defined by \[u_1 = 6 \text{ and } u_{n+1} = 6u_n - 5, \quad \text{for } n \geqslant 1.\]
Prove by induction that \(u_n = 5 \times 6^{n-1} + 1\), for \(n \geqslant 1\). (5)
| Scheme | Marks |
|---|---|
| At \(n = 1\), \(u_n = 5 \times 6^0 + 1 = 6\) and so result true for \(n = 1\) | B1 |
| Assume true for \(n = k\); \(u_k = 5 \times 6^{k-1} + 1\), and so \(u_{k+1} = 6(5 \times 6^{k-1} + 1) - 5\) | M1, A1 |
| \(\therefore u_{k+1} = 5 \times 6^k + 6 - 5 \qquad \therefore u_{k+1} = 5 \times 6^k + 1\) | A1 |
| and so result is true for \(n = k + 1\) and by induction true for \(n \geqslant 1\) | B1 |
| [5] |
Notes
6 and so result true for \(n = 1\) award B1
Sub \(u_k\) into \(u_{k+1}\) or M1 and A1 for correct expression on right hand of line 2
Second A1 for \(\therefore u_{k+1} = 5 \times 6^k + 1\)
‘Assume true for \(n = k\)’ and ‘so result is true for \(n = k + 1\)’ and correct solution for final B1