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,
$$\begin{gathered} P\left(n\right):\frac{n^7}{7}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{n^2}{2}-\frac{37}{210}n\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}. \\ \mathrm{Step}1: \\ P\left(1\right)=\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}=\frac{30+42+70+105-37}{210}=\frac{210}{210}=1 \\ Itis\mathrm{a}\mathrm{positive}\mathrm{integer}. \\ \mathrm{Thus},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ \mathrm{Then},\frac{m^7}{7}+\frac{m^5}{5}+\frac{m^3}{3}+\frac{m^2}{2}-\frac{37}{210}m\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}. \\ \mathrm{Let}\frac{m^7}{7}+\frac{m^5}{5}+\frac{m^3}{3}+\frac{m^2}{2}-\frac{37}{210}m=\lambda \mathrm{for}\mathrm{some}\lambda \in \mathrm{positive}N. \\ \mathrm{To}\mathrm{show}:P\left(m+1\right)\mathrm{is}a\mathrm{positive}\mathrm{integer}. \\ Now, \\ P(m+1)=\frac{{\left(m+1\right)}^7}{7}+\frac{{\left(m+1\right)}^5}{5}+\frac{{\left(m+1\right)}^3}{3}+\frac{{\left(m+1\right)}^2}{2}-\frac{37}{210}\left(m+1\right) \\ =\frac{1}{7}\left(m^7+7m^6+21m^5+35m^4+35m^3+21m^2+7m+1\right) \\ +\frac{1}{5}\left(m^5+5m^4+10m^3+10m^2+5m+1\right)+\frac{1}{3}\left(m^3+3m^2+3m+1\right)+\frac{1}{2}\left(m^2+2m+1\right)-\frac{37}{210}m-\frac{37}{210} \\ =\left[\frac{m^7}{7}+\frac{m^5}{5}+\frac{m^3}{3}+\frac{m^2}{2}-\frac{37}{210}m\right]+m^6+3m^5+6m^4+7m^3+6m^2+4m \\ =\lambda +m^6+3m^5+6m^4+7m^3+6m^2+4m \\ Itisa\mathrm{positive}\mathrm{integer},as\lambda \mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}. \\ \mathrm{Thus},P\left(m+1\right)\mathrm{is}\mathrm{true}, \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$