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Q.
For all positive integers $n, \frac{n^{7}}{7} +\frac{n^{5}}{5} +\frac{2n^{3}}{3} -\frac{n}{105}$ is a/an
Principle of Mathematical Induction
Solution:
Let $P\left(n\right)= \frac{n^{7}}{7} +\frac{n^{5}}{5} +\frac{2n^{3}}{3} -\frac{n}{105}$ is an integer.
For $n= 1$
$P\left(1\right) \frac{\left(1\right)^{7}}{7}+\frac{\left(1\right)^{5}}{5} +\frac{2\left(1\right)^{3}}{3}-\frac{\left(1\right)}{105}$
$=\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}$
$=\frac{ 15+21+70-1 }{105} =\frac{ 106-1}{105} $
$ = \frac{105}{105} = 1$
$\therefore P\left(1\right)$ is true.
Let $P\left(k\right)$ be true, then
$P\left(k\right) = \frac{k^{7}}{7} +\frac{k^{5}}{5}+\frac{2k^{3}}{3}-\frac{k}{105}$ is an integer
$\therefore \frac{k^{7}}{7} +\frac{k^{5}}{5}+\frac{2k^{3}}{3} -\frac{k}{105} = m, m\in Z\quad...\left(i\right)$
For $n = k + 1$, we have
$ P\left(k+1\right)=\frac{\left(k+1\right)^{7}}{7} +\frac{\left(k+1\right)^{5}}{5} +\frac{2\left(k+1\right)^{3}}{3} -\frac{\left(k+1\right)}{105}$
$=\frac{1}{7} \left(k^{7}+7k^{6}+21k^{5}+35k^{4}+35k^{3}+21k^{2}+7k+1\right)$
$+ \frac{1}{5} \left(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1\right)$
$+\frac{2}{3}\left(k^{3}+3k^{2}+3k+1\right) - \frac{k}{105}-\frac{1}{105}$
$ = \left(\frac{k^{7}}{7} +\frac{k^{5}}{5}+\frac{2k^{3}}{3}-\frac{k}{105}\right) + k^{6}+3k^{5}+6k^{4}$
$+7k^{3}+7k^{2}+4k+1 $
$= m+k^{6}+3k^{5}+6k^{4}+7k^{3}+7k^{2}+4k+1\quad$[Using $\left(i\right)$]
= an integer
Thus $P\left(k\right)$ is true $\Rightarrow P\left(k + 1\right)$ is true.
Hence by principle of mathematical induction, $P\left(n\right)$ is true
for all positive integers $n$.