Solve the following

Solution:

Let P(n) be the given statement.

Now,

$P(n): \frac{n^{11}}{11}+\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{62}{165} n$ is a positive integer for all $n \in N$.

Step 1:

$P(1)=\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165}=\frac{15+33+55+62}{165}=\frac{165}{165}=1$

It is certainly a positive integer.

Hence, $P(1)$ is true.

Step 2 :

Le $t P(m)$ be true.

Then, $\frac{m^{11}}{11}+\frac{m^{5}}{5}+\frac{m^{3}}{3}+\frac{62}{165} m$ is a positive integer.

Now, let $\frac{m^{11}}{11}+\frac{m^{5}}{5}+\frac{m^{3}}{3}+\frac{62}{165} m=\lambda$, where $\lambda \in N$ is a positive integer.

We have to show that $P(m+1)$ is true whenever $P(m)$ is true.

To prove : $\frac{(m+1)^{11}}{11}+\frac{(m+1)^{5}}{5}+\frac{(m+1)^{3}}{3}+\frac{62}{165}(m+1)$ is a positive integer.

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}+11 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{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}$$+3 m+\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{6}{105}$

$=\lambda+m^{10}+5 m^{9}+15 m^{8}+30 m^{7}+42 m^{6}+42 m^{5}+31 m^{4}+17 m^{3}+8 m^{2}+3 m+1$

It is a positive integer.

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

By the principle of mathematical induction, $P(n)$ is true for all $n \in N$.