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?

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

\[3k^2 + 3k + 6\]

?

\[3k(k+1) + 6\]

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.