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Question
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
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Solution
Let P(n) be the given statement.
Now,
\[P\left( n \right): {11}^{n + 2} + {12}^{2n + 1} \text{ is divisible by} 133 . \]
\[\text{ Step } 1: \]
\[P\left( 1 \right) = {11}^{1 + 2} + {12}^{2 + 1} = 1331 + 1728 = 3059 \]
\[\text{ It is divisible by } 133 . \]
\[\text{ Step2: }\]
\[\text{ Let } P\left( m \right)\text{ be divisible by } 133 . \]
\[Now, \]
\[ {11}^{m + 2} + {12}^{2m + 1}\text{ is divisible by } 133 . \]
\[\text{ Suppose: } \]
\[ {11}^{m + 2} + {12}^{2m + 1} = 133\lambda . . . (1)\]
\[\text{ We shall show that} P\left( m + 1 \right)\text{ is true whenever } P\left( m \right) \text{ is true .} \]
\[\text{ Now }, \]
\[P\left( m + 1 \right) = {11}^{m + 3} + {12}^{2m + 3} \]
\[ = {11}^{m + 2} . 11 + {12}^{2m + 1} . {12}^2 + 11 . {12}^{2m + 1} - 11 . {12}^{2m + 1} \]
\[ = 11\left( {11}^{m + 2} + {12}^{2m + 1} \right) + {12}^{2m + 1} \left( 144 - 11 \right)\]
\[ = 11 . 133\lambda + {12}^{2m + 1} . 133 \left[ From (1) \right]\]
\[ = 133\left( 11\lambda + {12}^{2m + 1} \right) \]
\[\text{ It is divisible by } 133 . \]
\[\text{ Thus, } P\left( m + 1 \right) \text{ is true .} \]
\[\text{ By the principle of mathematical induction, P(n ) is true for all } n \in N . \]
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