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Q.
For all $n \in N, the\, sum \,of\, \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is
Principle of Mathematical Induction
Solution:
Let the statement $P(n)$ be defined as
$P\left(n\right) :\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is a natural number for all $n \in N$. Step I : For $n = 1,$
$P\left(1\right) : \frac{1}{5}+\frac{1}{3}+\frac{7}{15}=1 \in N$
Hence, it is true for $n = 1$. Step II : Let it is true for $n = k$,
i,e. $\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7k}{15}=\lambda \in N \,...\left(i\right)$ Step III : For $n = k + 1$,
$\frac{\left(k+1\right)^{5}}{5}+\frac{\left(k+1\right)^{3}}{3}+\frac{7\left(k+1\right)}{15}$
$=\frac{1}{5}\left(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1\right) +\frac{1}{3}\left(k^{3}+3k^{2}+3k+1\right)+\frac{7}{15}k+\frac{7}{15}$
$=\left(\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7}{15}k\right)+\left(k^{4}+2k^{3}+3k^{2}+2k\right)+\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$
$=\lambda+k^{4}+2k^{3}+3k^{2}+2k+1$ [using equation (i)]
which is a natural number, since $\lambda k \in N.$
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 n.