<s>[INST] 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.
 [/INST] 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.
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