Standard +0.3 This is a straightforward proof by induction with a recurrence relation. The base case is trivial (u₁ = 2 = 5⁰ + 1), and the inductive step requires only direct substitution of the formula into the recurrence relation and basic algebraic manipulation. While it's a Further Maths topic, it follows the standard induction template with no conceptual surprises, making it slightly easier than average overall but typical for FP1 induction questions.
3. A sequence of numbers is defined by
$$\begin{aligned}
u _ { 1 } & = 2 \\
u _ { n + 1 } & = 5 u _ { n } - 4 , \quad n \geqslant 1 .
\end{aligned}$$
Prove by induction that, for \(n \in \mathbb { Z } ^ { + } , u _ { n } = 5 ^ { n - 1 } + 1\).
\(5(5^{k-1}+1)-4\) seen for M1; \(5^k+1\) or \(5^{(k+1)-1}+1\) for first A1
\(\therefore\) True for \(n=k+1\) if true for \(n=k\). True for \(n=1\), \(\therefore\) true for all \(n\).
A1 cso (4)
All three elements stated somewhere in solution for final A1
## Question 3:
For $n=1$: $u_1 = 2$, $u_1 = 5^0 + 1 = 2$ | B1 | Accept $u_1 = 1+1=2$
Assume true for $n=k$:
$u_{k+1} = 5u_k - 4 = 5(5^{k-1}+1)-4 = 5^k + 5 - 4 = 5^k + 1$ | M1 A1 | $5(5^{k-1}+1)-4$ seen for M1; $5^k+1$ or $5^{(k+1)-1}+1$ for first A1
$\therefore$ True for $n=k+1$ if true for $n=k$. True for $n=1$, $\therefore$ true for all $n$. | A1 cso (4) | All three elements stated somewhere in solution for final A1
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3. A sequence of numbers is defined by
$$\begin{aligned}
u _ { 1 } & = 2 \\
u _ { n + 1 } & = 5 u _ { n } - 4 , \quad n \geqslant 1 .
\end{aligned}$$
Prove by induction that, for $n \in \mathbb { Z } ^ { + } , u _ { n } = 5 ^ { n - 1 } + 1$.\\
\hfill \mbox{\textit{Edexcel FP1 2010 Q3 [4]}}