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Question
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
Solution
\[\text{ Given: A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 \text{ and } x_k = 4 + x_{k - 1} \text{ for all natural number } k . \]
\[\text{ Let } P\left( n \right): x_n = 5 + 4n \text{ for all } n \in N . \]
\[\text{ Step I: For } n = 0, \]
\[P\left( 0 \right): x_0 = 5 + 4 \times 0 = 5\]
\[\text{ So, it is true for } n = 0 . \]
\[\text{ Step II: For } n = k, \]
\[\text{ Let } P\left( k \right): x_k = 5 + 4 \text{ k be true for some k } \in N . \]
\[\text{ Step III: For } n = k + 1, \]
\[P\left( k + 1 \right): x_{k + 1} = 4 + x_{k + 1 - 1} \]
\[ = 4 + x_k \]
\[ = 4 + 5 + 4k\]
\[ = 5 + 4\left( k + 1 \right)\]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence } , x_n = 5 + 4\text{ n for all } n \in N .\]
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