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# Solution required

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

Note by Deepansh Jindal
1 year, 1 month ago

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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 $$n$$. · 1 year, 1 month ago

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