AS June 2020 Paper 1 Q7
7 Prove by induction that, for all integers \(n \geqslant 1\), the expression \(7^n - 3^n\) is divisible by 4 [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Shows that \(7^n - 3^n\) is divisible by 4 for \(n = 1\). | B1 | 1.1b |
| States the assumption that \(7^k - 3^k\) is divisible by 4 and considers \(7^{k+1} - 3^{k+1}\), by using \(7 \times 7^k\) or \(3 \times 3^k\). | M1 | 2.4 |
| Completes rigorous working to deduce that \(7^{k+1} - 3^{k+1}\) is divisible by 4. | R1 | 2.2a |
| Concludes a reasoned argument by stating that \(7^n - 3^n\) is divisible by 4 for \(n = 1\); that if \(7^k - 3^k\) is divisible by 4, then \(7^{k+1} - 3^{k+1}\) is divisible by 4 and hence, by induction, \(7^n - 3^n\) is divisible by 4 for \(n \geqslant 1\). | R1 | 2.1 |
| (4 marks) |
Typical solution
\[7^1 - 3^1 = 7 - 3 = 4\]Assume it is true for \(n = k\)
\[\therefore 7^k - 3^k = 4m\]where \(m\) is an integer
\[\begin{aligned}7^{k+1} - 3^{k+1} &= 7 \times 7^k - 3 \times 3^k \\ &= 7(4m + 3^k) - 3 \times 3^k \\ &= 28m + 4 \times 3^k \\ &= 4(7m + 3^k)\end{aligned}\]\(\therefore\) it is also true for \(n = k + 1\)
It is true for \(n = 1\). If it is true for \(n = k\) then it is true for \(n = k + 1\). Therefore, by induction, \(7^n - 3^n\) is divisible by 4 for all integers, \(n \geqslant 1\).