Further Maths - Induction for Divisibility
If you add a multiple of $4$ to something already divisible by $4$, what must be true about the answer?
It is also divisible by 4.
What is the two-step general technique used to show divisibility in induction?
- Assume $f(k)$ is divisible
- Show the difference between $f(k)$ and $f(k+1)$ is divisible.
How would you write the difference between $f(k)$ and $f(k+1)$?
If
\[f(k + 1) - f(k) = 4\times 3^{2k}\], what other statement could you write that shows clearly $f(k+1)$ is divisible by $4$??
\[f(k+1) = f(k) + 4\times 3^{2k}\]How could you simplify
\[3k^2 + 3k + 6\]??
\[3k(k+1) + 6\]#####
\[3k(k+1) + 6\]You’re trying to prove this statement is divisible by $6$. What new variable can you introduce??
\[2m = k(k+1),\quad m \in \mathbb{Z}^{+}\]#####
\[3k(k+1) + 6\]Why can you substitute $2m = k(k+1)$?? Because the product of any two consecutive integers must be even.
Substitute
\[2m = k(k+1)\]into
\[3k(k+1) + 6\]??
\[6m + 6\]If
\[f(k) = 2^{6k} + 3^{2k-2}\], how could you rewrite
\[63\times 2^{6k} + 8\times 3^{2k-2}\]in terms of $f(k)$??
\[63\times 2^{6k} + 63\times 3^{2k-2} - 55\times 3^{2k-2}\] \[63\times(2^{6k} + 3^{2k-2}) - 55\times3^{2k-2}\] \[63(f(k)) - 55\times3^{2k-2}\]If
\[f(k) = 2^{6k} + 3^{2k-2}\], what is the general technique (but not the process) to rewrite
\[63\times 2^{6k} + 8\times 3^{2k-2}\]in terms of $f(k)$?? Invent some extra terms that are cancelled out so it’s in the form you want.