9. A sequence of numbers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , u _ { 4 } , \ldots\) is defined by
$$u _ { n + 1 } = 4 u _ { n } + 2 , \quad u _ { 1 } = 2$$
Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\),
$$u _ { n } = \frac { 2 } { 3 } \left( 4 ^ { n } - 1 \right)$$
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Question 9:
\(u_{n+1} = 4u_n + 2\), \(u_1 = 2\) and \(u_n = \frac{2}{3}(4^n - 1)\)
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\(n=1\): \(u_1 = \frac{2}{3}(4^1-1) = \frac{2}{3}(3) = 2\), so \(u_n\) is true when \(n=1\) B1
Check that \(u_n = \frac{2}{3}(4^n-1)\) yields 2 when \(n=1\)
Assume for \(n=k\), \(u_k = \frac{2}{3}(4^k-1)\) is true for \(k \in \mathbb{Z}^+\). Then \(u_{k+1} = 4u_k + 2 = 4\left(\frac{2}{3}(4^k-1)\right)+2\) M1
Substituting \(u_k = \frac{2}{3}(4^k-1)\) into \(u_{n+1}=4u_n+2\)
\(= \frac{8}{3}(4)^k - \frac{8}{3} + 2\) M1
An attempt to multiply out brackets by 4 or \(\frac{8}{3}\)
\(= \frac{2}{3}(4)(4)^k - \frac{2}{3} = \frac{2}{3}4^{k+1} - \frac{2}{3} = \frac{2}{3}(4^{k+1}-1)\) A1
\(\frac{2}{3}(4^{k+1}-1)\)
Conclusion: true for \(n=k+1\), and since true for \(n=1\), true for all positive integers by induction A1
Require 'True when \(n=1\)', 'Assume true when \(n=k\)' and 'True when \(n=k+1\) then true for all \(n\) o.e.'
Total: 5 marks
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## Question 9:
$u_{n+1} = 4u_n + 2$, $u_1 = 2$ and $u_n = \frac{2}{3}(4^n - 1)$
| Working | Mark | Guidance |
|---------|------|----------|
| $n=1$: $u_1 = \frac{2}{3}(4^1-1) = \frac{2}{3}(3) = 2$, so $u_n$ is true when $n=1$ | B1 | Check that $u_n = \frac{2}{3}(4^n-1)$ yields 2 when $n=1$ |
| Assume for $n=k$, $u_k = \frac{2}{3}(4^k-1)$ is true for $k \in \mathbb{Z}^+$. Then $u_{k+1} = 4u_k + 2 = 4\left(\frac{2}{3}(4^k-1)\right)+2$ | M1 | Substituting $u_k = \frac{2}{3}(4^k-1)$ into $u_{n+1}=4u_n+2$ |
| $= \frac{8}{3}(4)^k - \frac{8}{3} + 2$ | M1 | An attempt to multiply out brackets by 4 or $\frac{8}{3}$ |
| $= \frac{2}{3}(4)(4)^k - \frac{2}{3} = \frac{2}{3}4^{k+1} - \frac{2}{3} = \frac{2}{3}(4^{k+1}-1)$ | A1 | $\frac{2}{3}(4^{k+1}-1)$ |
| Conclusion: true for $n=k+1$, and since true for $n=1$, true for all positive integers by induction | A1 | Require 'True when $n=1$', 'Assume true when $n=k$' and 'True when $n=k+1$ then true for all $n$ o.e.' |
**Total: 5 marks**
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9. A sequence of numbers $u _ { 1 } , u _ { 2 } , u _ { 3 } , u _ { 4 } , \ldots$ is defined by
$$u _ { n + 1 } = 4 u _ { n } + 2 , \quad u _ { 1 } = 2$$
Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
$$u _ { n } = \frac { 2 } { 3 } \left( 4 ^ { n } - 1 \right)$$
\hfill \mbox{\textit{Edexcel FP1 2011 Q9 [5]}}