Prove by the principle of mathematical induction that `(n^5)/5+(n^3)/3+(7n)/(15)` is a natural number for all `n in Ndot`

Explanation

Let P(n): `(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)` is a natural number.
When n=1, `(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)=(1^(5))/(5)+(1^(3))/(3)+(7)/(15)=(1)/(5)+(1)/(3)+(7)/(15)=1`, which is a natual number
Hence, P(1) is true……(A)
Let P(m) be true
`Rightarrow (m^(5))/(5)+(m^(3))/(3)+(7m)/(15)` is a natural number .....(i)
To prove P(m+1) is true i.e,
`(m+1)^(5)/(5)+(m+1)^(3)/(3)+(7(m+1))/(15)` is natural number .....(ii)
Expanding (ii), we get
`(1)/(5)(m^(5)+5m^4+10m^(3)+10m^(2)+5m+1)+(1)/(3)(m^(3)+m^(2)+3m+1)+(7)/(15)(m+1)`
`=(m^(5))/(5)+(m^(3))/(3)+(7m)/(15)+(m^(4)+2m^(3)+3m^(2)+2m)+(1)/(5)+(1)/(3)+(7)/(15)`
`=(m^(5))/(5)+(m^(3))/(3)+(7m)/(15)+(m^(4)+2m^(3)+2m+1)`
=a natural number `[therefore m^(5)/(5)+(m^(3))/(3)+(7m)/(15)"is a natural number from (i) "]`
Hence P(m+1) is true whenever P(m) is true .....(B)
From (A) and (B) it follows that P(n) is true for all natural number n.