Solution required

Prove that for any natural number $n$ by mathematical induction that $4^{n} +15n -1$ is divisible by 9.

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Comments

Let $P_n$ be the proposition $9~|~4^{n}+15n-1$ for any natural number $n$ .

First, consider the base case where $n = 1$ . Observe that

$4^{1} + 15 (1) - 1 = 4 + 15 - 1 = 18$

and that $9$ divides $18$ . Then $P_1$ is true.

Now, assume $P_{k}$ is true for some $k \in \text{domain},$ then $9 ~|~4^{n} + 15n - 1$ .

We realize that $4^{k+1}=4\times 4^{k}$ . So

$\begin{array} { l l } 9~|~ 4^{k+1}+15k\times 4 - 1\times 4\\ 9~|~ 4^{k+1}+15k + 15 + (3 \times 15k - 15) -1 + (-1\times 3)\\ 9~|~ 4^{k+1}+15(k+1) -1 + (45k -18)\\ 9~|~ 4^{(k+1)}+15(k+1) -1. \end{array}$

Hence $P_{k}$ true $\Rightarrow P_{k+1}$ true.

By mathematical induction, since the base case where $n=1$ is true and $P_{k}$ is true, $P_{k+1}$ true. Therefore, $P_{n}$ is true for all natural numbers $n$ .

Francisco Rodríguez - 4 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$