Show that `(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)` is a natural number, for all n `inN`

Explanation

Let `P(n) : (n^(5))/(5) + (n^(3))/(3) + (7n)/(15)` is a natural number, for all `n in N`.
`P(1) : (1^(5))/(5) + (1^(3))/(3) + (7(1))/(15) = (3+5+7)/(15) = 1515 = 1`, which is a natural number.
Hence, `P(10` is true.
Let `P(n)` be true for some `n = k`.
Then `(k^(5))/(5) + (k^(3))/(3) + (7k)/(15)` is natural number.
Now, `((k+1)^(5))/(5) + ((k+1)^(3))/(3) (7(k+1))/(15)`
`= (k^(5) + 5k^(4) + 10 k^(3) + 10 k^(2) + 5k +1)/(5)`
` + (k^(3) + 1 + 3k^(2) + 3k)/(3) + (7k + 7)/(15)`
`= (k^(5))/(5) + (k^(3))/(3) + (7k)/(15) + k^(4) + 2k^(3) + 3k^(2) + 2k + 1`
`= P(k) + k^(4) + 2k^(3) + 2k^(2) + 1`
`=` Natural number
So, by the principle of mathematical induction, `P(n)` is true for any natural n.