Edexcel A-Level Mathematics AS Fp1 8 1 Proof By Mathematical Induction Questions

Let
$$A=\left(\text{row }1:5,\ -1;\ \text{row }2:4,\ 1\right).$$
For n=1, the right-hand side gives A. B1

Assume true for n=k:
$$A^k=3^{k-1}\left(\text{row }1:2k+3,\ -k;\ \text{row }2:4k,\ 3-2k\right).$$
Then
$$A^{k+1}=3^{k-1} \left(\text{row }1:2k+3,\ -k;\ \text{row }2:4k,\ 3-2k\right)A.$$$$=3^{k-1}\left(\text{row }1:6k+15,\ -3k-3;\ \text{row }2:12k+12,\ -6k+3\right).$$$$=3^k\left(\text{row }1:2(k+1)+3,\ -(k+1);\ \text{row }2:4(k+1),\ 3-2(k+1)\right).$$
Thus true for n=k+1. Since true for n=1, the result is true for all $n\in\mathbb Z^+$. A1cso

$\mathrm{f}(1)=8^{3}+6=518=74 \times 7$ (so true for n=1 )
Assume true for n=k so that $8^{2 k+1}+6^{2 k-1}$ is divisible by 7

$\mathrm{f}(k+1)=8^{2 k+3}+6^{2 k+1}$$=64 \times\left(8^{2 k+1}+6^{2 k-1}\right)+\ldots$ or $36 \times\left(8^{2 k+1}+6^{2 k-1}\right)+\ldots$$=64 \times\left(8^{2 k+1}+6^{2 k-1}\right)-28 \times 6^{2 k-1}$ or $36 \times\left(8^{2 k+1}+6^{2 k-1}\right)+28 \times 8^{2 k+1}$
So if the result is true for n=k then it is true for n=k+1. As the result has
been shown to be true for n=1, then the result is true for all n.

A1cso

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