| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Topic | Proof by induction |
| Type | Prove divisibility |
| Difficulty | Standard +0.3 This is a standard proof by induction with divisibility, requiring verification of the base case and algebraic manipulation in the inductive step to factor out 21. The arithmetic is straightforward (showing 4^(k+2) + 5^(2k+1) = 4·4^(k+1) + 25·5^(2k-1) and using the inductive hypothesis), making this slightly easier than average but still requiring proper proof technique. |
| Spec | 4.01a Mathematical induction: construct proofs |
Prove by induction that for $n \in \mathbb{Z}^+$
$$f(n) = 4^{n+1} + 5^{2n-1}$$
is divisible by 21 [6]
\hfill \mbox{\textit{SPS SPS FM Pure 2022 Q9 [6]}}