Question

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

Answer 1

$P\left(n\right):\frac{n^{11}}{11}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{n^{62}}{165}$ n is a positive integer

For n = 1

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

Which is a positive integer

Let P(n) is true for n = k, so

$\frac{k^{11}}{11}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{62k}{165}$ is a positive integer

$\frac{k^{11}}{11}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{62k}{165}=\lambda$ ......(i)

For n = k + 1

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

$=\frac{1}{11}[k^{11}+11k^{10}+55k^9+165k^8+330k^7+462k^6+462k^5+330k^4+165k^3+55k^2+11k+1]+\frac{1}{5}[k^5+5k^4+10k^3+10k^2+5k+1]+\frac{1}{3}[k^3+3k^2+3k+1]+\frac{62}{165}[k+1]$

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

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

= An integer

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI.