Standard +0.3 This is a straightforward proof by induction with a given formula for a recurrence relation. The base case is simple arithmetic, and the inductive step requires standard algebraic manipulation of the recurrence relation. While it's a Further Maths topic, the technique is mechanical and well-practiced, making it slightly easier than average overall.
3 A sequence is defined by \(a _ { 1 } = 6\) and \(a _ { n + 1 } = 5 a _ { n } - 2\) for \(n \geqslant 1\).
Prove by induction that for all integers \(n \geqslant 1 , a _ { n } = \frac { 11 \times 5 ^ { n - 1 } + 1 } { 2 }\).
Use of both recurrence relation for \(a_{k+1}\) in terms of \(a_k\) and also inductive hypothesis in attempt to derive required form for \(a_{k+1}\)
So if true for \(n = k\) then true also for \(n = k+1\). But since it is true for \(n = 1\) then it must be true for all integers \(n \geq 1\).
E1 [5]
Base case, $n = 1$, $a_1 = \frac{11 \times 5^0 + 1}{2} = 6$ | B1 [5] |
Assume true for $n = k$: $\Rightarrow a_k = \frac{11 \times 5^{k-1} + 1}{2}$ | M1 [5] | Statement of inductive hypothesis
Then $a_{k+1} = 5a_k - 2 = 5(\frac{11 \times 5^{k-1} + 1}{2}) - 2 = (\frac{11 \times 5^k + 5}{2}) - (\frac{11 \times 5^k + 1 + 4}{2}) - 2 = (\frac{11 \times 5^k + 1}{2}) + 2 - 2 = \frac{11 \times 5^{(k+1)} + 1}{2}$ | M1, A1 [5] | Use of both recurrence relation for $a_{k+1}$ in terms of $a_k$ and also inductive hypothesis in attempt to derive required form for $a_{k+1}$
So if true for $n = k$ then true also for $n = k+1$. But since it is true for $n = 1$ then it must be true for all integers $n \geq 1$. | E1 [5]
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3 A sequence is defined by $a _ { 1 } = 6$ and $a _ { n + 1 } = 5 a _ { n } - 2$ for $n \geqslant 1$.\\
Prove by induction that for all integers $n \geqslant 1 , a _ { n } = \frac { 11 \times 5 ^ { n - 1 } + 1 } { 2 }$.
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q3 [5]}}