Standard +0.3 This is a straightforward proof by induction for divisibility. The base case is trivial (9^1 + 15 = 24), and the inductive step requires only basic algebraic manipulation: 9^(k+1) + 15 = 9(9^k + 15) - 120, both terms divisible by 8. Standard template application with no conceptual challenges, making it slightly easier than average.
Each of the two terms are multiples of 8 so therefore is the left hand side.
A1
So, if proposition is true for \(n = k\), it's also true for \(n = k+1\). Since we have shown it's true for \(n = 1\), by mathematical induction, it's true for all positive integers \(n\).
A1
cso
When $n = 1$, $9^1 + 15 = 24$ which is a multiple of 8. Therefore, proposition is true for $n = 1. | B1 |
Assume the proposition is true for $n = k$, i.e. $9^k + 15$ is a multiple of 8 or $9^k + 15 = 8N$ | M1 |
Consider $n = k+1$: $9^{k+1} + 15 = 9(9^k) + 15 = 9(8N - 15) + 15 = 72N - 120$ | M1, A1, A1 |
Each of the two terms are multiples of 8 so therefore is the left hand side. | A1 |
So, if proposition is true for $n = k$, it's also true for $n = k+1$. Since we have shown it's true for $n = 1$, by mathematical induction, it's true for all positive integers $n$. | A1 | cso
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4. Prove, by mathematical induction, that $9 ^ { n } + 15$ is a multiple of 8 for all positive integers $n$.\\
\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q4 [7]}}