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 $f(k+1)$??
$$ f(k+1) - f(k) $$
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.
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date: 2021-01-05 11:18
tags:
- '@?further-maths'
- '@?induction'
- '@?public'
title: Further Maths - Induction for Divisibility