Use the method of mathematical induction to prove that \({5^{2n}} - 24n - 1\) is divisible by 576 for \(n \in {\mathbb{Z}^ + }\).

\({\text{P}}(n):f(n) = {5^{2n}} - 24n - 1\) is divisible by 576 for \(n \in {\mathbb{Z}^ + }\)

for \(n = 1,{\text{ }}f(1) = {5^2} - 24 - 1 = 0\)

Zero is divisible by 576, (as every non-zero number divides zero), and so P(1) is true.     R1

Note: Award R0 for P(1) = 0 shown and zero is divisible by 576 not specified.

Note: Ignore P(2) = 576 if P(1) = 0 is shown and zero is divisible by 576 is specified.

Assume \({\text{P}}(k)\) is true for some \(k{\text{ }}( \Rightarrow f(k) = N \times 576)\).     M1

Note: Do not award M1 for statements such as “let n = k”.

consider \({\text{P}}(k + 1):f(k + 1) = {5^{2(k + 1)}} - 24(k + 1) - 1\)     M1

\( = 25 \times {5^{2k}} - 24k - 25\)     A1

EITHER

\( = 25 \times (24k + 1 + N \times 576) - 24k - 25\)     A1

\( = 576k + 25 \times 576N\) which is a multiple of 576     A1

OR

\( = 25 \times {5^{2k}} - 600k - 25 + 600k - 24k\)     A1

\( = 25({5^{2k}} - 24k - 1) + 576k\) (or equivalent) which is a multiple of 576     A1

THEN

P(1) is true and \({\text{P}}(k)\) true \( \Rightarrow {\text{P}}(k + 1)\) true, so \({\text{P}}(n)\) is true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Award R1 only if at least four prior marks have been awarded.

[7 marks]

This proof by mathematical induction challenged most candidates. While most candidates were able to show that P(1) = 0, a significant number did not state that zero is divisible by 576. A few candidates started their proof by looking at P(2). It was pleasing to see that the inductive step was reasonably well done by most candidates. However many candidates committed simple algebraic errors. The most common error was to state that \({5^{2(k + 1)}} = 5{(5)^{2k}}\). The concluding statement often omitted the required implication statement and also often omitted that P(1) was found to be true.