Use the Principle of Mathematical Induction to show that $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is a natural number for all $n \in N$.
(N/A) $P(n): \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is a natural number,$n \in N$.
For $n=1$,$P(1) = \frac{1^{5}}{5} + \frac{1^{3}}{3} + \frac{7(1)}{15} = \frac{3+5+7}{15} = \frac{15}{15} = 1$,which is a natural number.
Therefore,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,$\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7k}{15} = m$,where $m \in N$.
For $n=k+1$,we have $P(k+1) = \frac{(k+1)^{5}}{5} + \frac{(k+1)^{3}}{3} + \frac{7(k+1)}{15}$.
Expanding the terms: $P(k+1) = \frac{k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1}{5} + \frac{k^{3}+3k^{2}+3k+1}{3} + \frac{7k+7}{15}$.
Rearranging: $P(k+1) = (\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7k}{15}) + (k^{4} + 2k^{3} + 2k^{2} + k) + (k^{2} + k) + (\frac{1}{5} + \frac{1}{3} + \frac{7}{15})$.
$P(k+1) = m + k^{4} + 2k^{3} + 3k^{2} + 2k + (\frac{3+5+7}{15}) = m + k^{4} + 2k^{3} + 3k^{2} + 2k + 1$.
Since $m, k \in N$,$P(k+1)$ is also a natural number.
Thus,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.
For $n=1$,$P(1) = \frac{1^{5}}{5} + \frac{1^{3}}{3} + \frac{7(1)}{15} = \frac{3+5+7}{15} = \frac{15}{15} = 1$,which is a natural number.
Therefore,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,$\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7k}{15} = m$,where $m \in N$.
For $n=k+1$,we have $P(k+1) = \frac{(k+1)^{5}}{5} + \frac{(k+1)^{3}}{3} + \frac{7(k+1)}{15}$.
Expanding the terms: $P(k+1) = \frac{k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1}{5} + \frac{k^{3}+3k^{2}+3k+1}{3} + \frac{7k+7}{15}$.
Rearranging: $P(k+1) = (\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7k}{15}) + (k^{4} + 2k^{3} + 2k^{2} + k) + (k^{2} + k) + (\frac{1}{5} + \frac{1}{3} + \frac{7}{15})$.
$P(k+1) = m + k^{4} + 2k^{3} + 3k^{2} + 2k + (\frac{3+5+7}{15}) = m + k^{4} + 2k^{3} + 3k^{2} + 2k + 1$.
Since $m, k \in N$,$P(k+1)$ is also a natural number.
Thus,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.
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