Solve the following
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 $n \in N$.

Solution:

Let P(n) be the given statement.

Now,

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

Step 1:

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

It is a positive integer.

Thus, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

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

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

To show : $P(m+1)$ is $a$ positive integer.

Now,

$P(m+1)=\frac{(m+1)^{7}}{7}+\frac{(m+1)^{5}}{5}+\frac{(m+1)^{3}}{3}+\frac{(m+1)^{2}}{2}-\frac{37}{210}(m+1)$

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

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

$-\frac{37}{210}$

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

$=\lambda+m^{6}+3 m^{5}+6 m^{4}+7 m^{3}+6 m^{2}+4 m$

It is a positive integer, $a s \lambda$ is a positive integer.

Thus, $P(m+1)$ is true,

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.