proof using the mathematical induction Prove $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer for every positive integer k.
In proving for (n+1) integer,the expression is integer,I found $(n+1)^7$ term.I use binomial theorem to expand but finally it won't work(some term become integer While some other remaining as rational)
Any hint to solve the problem is appreciated.Thanks.

Based on other similar questions, you might think to first prove that
$$ \frac{k^7 - k}{7} $$
is always an integer. Once you have done so, you can simplify
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=
\frac{k^7 - k}{7} 
+\frac{k^5}{5}+\frac{2k^3}{3}+\frac{2k}{15}
$$ 
You can do the same with $5$ and $3$, leaving you with
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=
\frac{k^7 - k}{7} + \frac{k^5 - k}{5} + 2 \frac{k^3 - k}{3}   + k
$$
Another path to the same approach might be to use a variant of partial fractions to separate the denominator into its prime components:
$$ \frac{1}{105} = \frac{1}{7} + \frac{1}{5} + \frac{2}{3} - 1$$
and then group like denominators together.
