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Question
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Solution
Let P(n) be the given statement.
Now,
\[P(n): \frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n \text{ is a positive integer for all } n \in N . \]
\[\text{ Step } 1: \]
\[P(1) = \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{62}{165} = \frac{15 + 33 + 55 + 62}{165} = \frac{165}{165} = 1 \]
\[\text{ It is certainly a positive integer } . \]
\[\text{ Hence, P(1) is true .} \]
\[\text{ Step2: } \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then, } \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m \text{ is a positive integer . } \]
\[\text{ Now, let } \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m = \lambda, \text{ where } \lambda \in N\text{ is a positive integer . } \]
\[\text{ We have to show that P(m + 1) is true whenever P(m) is true } . \]
\[\text{ To prove: } \frac{(m + 1 )^{11}}{11} + \frac{(m + 1 )^5}{5} + \frac{(m + 1 )^3}{3} + \frac{62}{165}(m + 1)\text{ is a positive integer .} \]
\[\text{ Now, } \]
\[\frac{(m + 1 )^{11}}{11} + \frac{(m + 1 )^5}{5} + \frac{(m + 1 )^3}{3} + \frac{62}{165}(m + 1)\]
\[ = \frac{1}{11}\left( m^{11} + 11 m^{10} + 55 m^9 + 165 m^8 + 330 m^7 + 462 m^6 + 462 m^5 + 330 m^4 + 165 m^3 + 55 m^2 + 11m + 1 \right)\]
\[ + \frac{1}{5}\left( m^5 + 5 m^4 + 10 m^3 + 10 m^2 + 5m + 1 \right) + \frac{1}{3}\left( m^3 + 3 m^2 + 3m + 1 \right)\]
\[ + \frac{62}{165}m + \frac{62}{165}\]
\[ = \left[ \frac{m^{11}}{11} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{62}{165}m \right] + m^{10} + 5 m^9 + 15 m^8 + 30 m^7 + 42 m^6 + 42 m^5 + 31 m^4 + 17 m^3 + 8 m^2 + 3m + \frac{1}{11} + \frac{1}{5} + \frac{1}{3} + \frac{6}{105}\]
\[ = \lambda + m^{10} + 5 m^9 + 15 m^8 + 30 m^7 + 42 m^6 + 42 m^5 + 31 m^4 + 17 m^3 + 8 m^2 + 3m + 1\]
\[\text{ It is a positive integer } . \]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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