Moderate -0.3 This is a straightforward proof by induction with a simple divisibility statement. The base case is trivial (n=1 gives 7+1=8), and the inductive step requires only basic algebraic manipulation to show 7^(k+1) + 3^k = 7(7^k + 3^(k-1)) - 4ยท3^(k-1), making the factor of 4 evident. While it's a Further Maths topic, it's a standard textbook exercise requiring no novel insight, making it slightly easier than average overall.
Where \(\lambda'\) is an integer because \(\lambda\) is and so rhs is a multiple of 4.
A1
So if true for \(n = k\), true also for \(n = k+1\). But it is true for \(n = 1\). Therefore true for all positive integers.
A1
Base case: For $n = 1$, $7^n + 3^{n-1} = 7+1 = 8$ which is a multiple of 4. | B1 |
Assume that $f(k) = 7^k + 3^{k-1} = 4\lambda$ for some integer, $\lambda$ | M1 |
$f(k+1) = 7^{k+1} + 3^k = 7.7^k + (7-4)3^{k-1} = 7f(k) - 4.3^{k-1} = 7.4\lambda - 4.3^{k-1} = 4(7\lambda - 3^{k-1}) = 4\lambda'$ | M1 | Alternatively: $f(k+1) = 7^{k+1} + 3^k = 7.7^k + 3.3^{k-1} = 7.7^k + 3.3^{k-1} = 7(4k - 3^{k-1}) + 3.3^{k-1} = ...$
Where $\lambda'$ is an integer because $\lambda$ is and so rhs is a multiple of 4. | A1 |
So if true for $n = k$, true also for $n = k+1$. But it is true for $n = 1$. Therefore true for all positive integers. | A1 |
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6 Prove by induction that, for all positive integers $n , 7 ^ { n } + 3 ^ { n - 1 }$ is a multiple of 4 .
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q6 [5]}}