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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, 10 months 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 \(n\).

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TopNewestLet \(P_n \) 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 \(n\).

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