Question

$$\frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{62}{165}n$$ is a positive integer for all n ∈ N. 

 

Answer 1

Let P(n) be the given statement.
Now, 

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

$$\text{ Step }1:$$

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

$$\text{ It is certainly a positive integer }.$$

$$\text{ Hence, P(1) is true .}$$

$$\text{ Step2: }$$

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

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

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

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

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

$$\text{ Now, }$$

$$\frac{(m+1{)}^{11}}{11}+\frac{(m+1{)}^5}{5}+\frac{(m+1{)}^3}{3}+\frac{62}{165}(m+1)$$

$$=\frac{1}{11}(m^{11}+11m^{10}+55m^9+165m^8+330m^7+462m^6+462m^5+330m^4+165m^3+55m^2+11m+1)$$

$$+\frac{1}{5}(m^5+5m^4+10m^3+10m^2+5m+1)+\frac{1}{3}(m^3+3m^2+3m+1)$$

$$+\frac{62}{165}m+\frac{62}{165}$$

$$=[\frac{m^{11}}{11}+\frac{m^5}{5}+\frac{m^3}{3}+\frac{62}{165}m]+m^{10}+5m^9+15m^8+30m^7+42m^6+42m^5+31m^4+17m^3+8m^2+3m+\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{6}{105}$$

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

$$\text{ It is a positive integer }.$$

$$\text{ Thus, P(m + 1) is true }.$$

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