Question

\[\frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n\] is a positive integer for all n ∈ N.  

 

Answer 1

Let P(n) be the given statement.
Now,

\[P(n): \frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n \text{ is a positive integer .}  \]

\[\text{ Step 1:}  \]

\[P(1) = \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210} = \frac{30 + 42 + 70 + 105 - 37}{210} = \frac{210}{210} = 1 \]

\[\text{ It is a positive integer .}  \]

\[\text{ Thus, P(1) is true }  . \]

\[\text{ Step } 2: \]

\[\text{ Let P(m) be true } . \]

\[\text{ Then } , \frac{m^7}{7} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{m^2}{2} - \frac{37}{210}m \text{ is a positive integer } . \]

\[Let \frac{m^7}{7} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{m^2}{2} - \frac{37}{210}m = \lambda \text{ for some } \lambda \in \text{ positive } N . \]

\[\text{ To show: } P\left( m + 1 \right) \text { is a positive integer } . \]

\[\text{ Now } , \]

\[P(m + 1) = \frac{\left( m + 1 \right)^7}{7} + \frac{\left( m + 1 \right)^5}{5} + \frac{\left( m + 1 \right)^3}{3} + \frac{\left( m + 1 \right)^2}{2} - \frac{37}{210}\left( m + 1 \right)\]

\[ = \frac{1}{7}\left( m^7 + 7 m^6 + 21 m^5 + 35 m^4 + 35 m^3 + 21 m^2 + 7m + 1 \right)\]

\[ + \frac{1}{5}\left( m^5 + 5 m^4 + 10 m^3 + 10 m^2 + 5m + 1 \right) + \frac{1}{3}\left( m^3 + 3 m^2 + 3m + 1 \right) + \frac{1}{2}\left( m^2 + 2m + 1 \right) - \frac{37}{210}m - \frac{37}{210} \]

\[ = \left[ \frac{m^7}{7} + \frac{m^5}{5} + \frac{m^3}{3} + \frac{m^2}{2} - \frac{37}{210}m \right] + m^6 + 3 m^5 + 6 m^4 + 7 m^3 + 6 m^2 + 4m\]

\[ = \lambda + m^6 + 3 m^5 + 6 m^4 + 7 m^3 + 6 m^2 + 4m\]

\[\text{ It is a positive integer, as \lambda is a positive integer } . \]

\[\text{ Thus } , P\left( m + 1 \right) \text{ is true } , \]

\[\text{ By the principle of mathematical induction, P(n) is true for all }  n \in N . \]