How to use Induction with Sequences? I have posted this similar question here, but with no hopes. I would just like to know:


Most of the solution I have no issue with. Look at where they say:
"Choose a representation $(n - 3^m)/2 = s_1 + ... + s_k$ in the desired form.
But to do that first, they applied induction to the set:
$$A = \{1, 2, 3, ..., n-1\}$$
You have to ensure,
$$\frac{n - 3^m}{2} \le n-1$$ 
How do you show that? 
Lets, consider the case, $n=5$. It follows,
$$3^1 < 5 < 3^2$$, Hence, $m = 1$ which gives:
$$\frac{5 - 3}{2} \le 4 \implies 1 < 4 \checkmark$$
But the question is how to prove it? 
Suppose:
$$\frac{n - 3^m}{2} \le n-1$$
I am to prove:
$$\frac{n + 1 - 3^m}{2} \le n$$
Begin with the hypothesis,
$$\frac{n - 3^m}{2} \le n-1$$
$$\frac{n - 3^m}{2} + \frac{1}{2} \le n- \frac{1}{2} < n$$
Since $n \in \mathbb{N}$
I suppose the statement is proved. Can you check it?

Recall that $m=\lfloor\log_3n\rfloor$. This means that $m\le\log_3n<m+1$ and hence that $3^m\le n<3^{m+1}$. It follows that $n-3^m<3^{m+1}-3^m$ and hence that
$$\frac{n-3^m}2<\frac{3^{m+1}-3^m}2=\frac{3^m(3-1)}2=3^m\le n\;.$$
That is,
$$\frac{n-3^m}2<n\;.\tag{1}$$
Since we’re assuming in this case that $n$ is odd, we know that $\frac{n-3^m}2$ is an integer, $(1)$ implies that
$$\frac{n-3^m}2\le n-1\;.$$
