Question

Prove that $\frac{n^7}{7}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{n^2}{2}-\frac{37}{210}$ n is a positive integer for all $nϵN$.

Answer 1

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

For n = 1

$\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$

$=\frac{30+42+70+105-37}{210}=\frac{247-37}{210}$

It is a positive integer

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

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

$\frac{k^7}{7}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{k^2}{2}-\frac{37}{210}$ k is positive integer

$\frac{k^7}{7}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{k^2}{2}-\frac{37}{210}k=\lambda$

For n = k + 1,

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

$=\frac{1}{7}[k^7+7k^6+21k^5+35k^4+35k^3+21k^2+7k+1]+\frac{1}{5}[k^5+5k^4+10k^3+10k^2+5k+1]+\frac{1}{3}[5^3+3k62+3k+1]+\frac{1}{2}[k^2+2k+1]-\frac{37k}{210}-\frac{37}{210}=[\frac{k^7}{7}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{k^2}{2}-\frac{37k}{210}]+[k^6+3k^5+5k^4+5k^3+3k^2+k+\frac{1}{7}+k^4+2k^3+2k^2+\frac{1}{5}+k^2+k+\frac{1}{3}+k+\frac{1}{2}-\frac{37}{210}]$

$=\lambda +k^6+3k^5+6k^4+7k^3+6k^2+3k+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$

$=\lambda +k^6+3k^5+6k^4+7k^3+6k^2+3k+1$

= Positive integer

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

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