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Question
\[\text{ A sequence } a_1 , a_2 , a_3 , . . . \text{ is defined by letting } a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
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Solution
\[\text{ Let } P\left( n \right): a_n = 3 \cdot 7^{n - 1} \text{ for all n } \in N . \]
\[\text{ Step I: For n } = 1, \]
\[P\left( 1 \right): \]
\[ a_1 = 3 \cdot 7^{1 - 1} = 3 \cdot 1 = 3\]
\[\text{ So, it is true for } n = 1 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } P\left( k \right): a_k = 3 \cdot 7^{k - 1}\text{ be true for some } k \in N \text{ and } k \geq 2 . \]
\[\text{ Step III: For n } = k + 1, \]
\[ a_{k + 1} = 7 a_k \]
\[ = 7 \cdot 3 \cdot 7^{k - 1} \left( \text{ Using step } II \right)\]
\[ = 3 \cdot 7^{k - 1 + 1} \]
\[ = 3 \cdot 7^\left( k + 1 \right) - 1 \]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence } , a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]
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