For all $n \in N$, the sum of $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7 n}{15}$ is

Let the statement $\mathrm{P}(\mathrm{n})$ be defined as $P(n): \frac{n^5}{5}+\frac{n^3}{3}+\frac{7 n}{15}$ is a natural number for all $n \in N$ :
Step I : For $\mathrm{n}=1, \mathrm{P}(1): \frac{1}{5}+\frac{1}{3}+\frac{7}{15}=1 \in \mathrm{N}$
Hence, it is true for $\mathrm{n}=1$.
Step II : Let it is true for $n=k$,
i.e. $\frac{k^5}{5}+\frac{k^3}{3}+\frac{7 k}{15}=\lambda \in N$ $\ldots$ (i)
Step III : For $\mathrm{n}=\mathrm{k}+1$,
$\frac{(\mathrm{k}+1)^5}{5}+\frac{(\mathrm{k}+1)^3}{3}+\frac{7(\mathrm{k}+1)}{15}$
$=\frac{1}{5}\left(k^5+5 k^4+10 k^3+10 k^2+3 k+1\right)$ $+\frac{1}{3}\left(k^3+3 k^2+3 k+1\right)+\frac{7}{15} k+\frac{7}{15}$
$=\left(\frac{k^5}{5}+\frac{k^3}{3}+\frac{7}{15} k\right)+\left(k^4+2 k^3+3 k^2+2 k\right)$ $+\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$
$=\lambda+\mathrm{k}^4+2 \mathrm{k}^3+3 \mathrm{k}^2+2 \mathrm{k}+1$ [using equation (i)]
which is a natural number, since $\lambda \mathrm{k} \in \mathrm{N}$. Therefore, $P(k+1)$ is true, when $P(k)$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $\mathrm{n}$