Standard +0.3 This is a standard proof by induction for a recurrence relation formula. It requires verifying the base case (n=1), assuming the formula holds for n=k, then proving it for n=k+1 by substituting into the recurrence relation. The algebra is straightforward (substituting and simplifying 6(5×6^(k-1)+1)-5). This is a textbook FP1 induction question with no novel insight required, making it slightly easier than average overall but typical for Further Maths.
6. A series of positive integers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by
$$u _ { 1 } = 6 \text { and } u _ { n + 1 } = 6 u _ { n } - 5 \text {, for } n \geqslant 1 \text {. }$$
Prove by induction that \(u _ { n } = 5 \times 6 ^ { n - 1 } + 1\), for \(n \geqslant 1\).
Result true for \(n=k+1\) and by induction true for \(n\geq1\)
B1
'Assume true for \(n=k\)' and 'result true for \(n=k+1\)' and correct solution for final B1
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $n=1$, $u_n = 5\times6^0+1 = 6$ and so result true for $n=1$ | B1 | 6 and so result true for $n=1$ |
| Assume true for $n=k$: $u_k = 5\times6^{k-1}+1$, and so $u_{k+1} = 6(5\times6^{k-1}+1)-5$ | M1, A1 | Sub $u_k$ into $u_{k+1}$ for M1; correct expression on RHS for A1 |
| $\therefore u_{k+1} = 5\times6^k+6-5 \therefore u_{k+1} = 5\times6^k+1$ | A1 | Second A1 for $u_{k+1} = 5\times6^k+1$ |
| Result true for $n=k+1$ and by induction true for $n\geq1$ | B1 | 'Assume true for $n=k$' and 'result true for $n=k+1$' and correct solution for final B1 |
6. A series of positive integers $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
$$u _ { 1 } = 6 \text { and } u _ { n + 1 } = 6 u _ { n } - 5 \text {, for } n \geqslant 1 \text {. }$$
Prove by induction that $u _ { n } = 5 \times 6 ^ { n - 1 } + 1$, for $n \geqslant 1$.\\
\hfill \mbox{\textit{Edexcel FP1 2009 Q6 [5]}}