Question

Prove that $n^3+3n^2+5n+3$ is divisible by $3$ for any natural $n$.

Answer 1

$P\left(n\right):n^3+{3n}^2+5n+3$ is divisible by 3

$\therefore P\left(1\right):1^3+{3\times 1}^2+5\times 1+3$

$=1+3+5+3=12$ which is divisible by 3

$\therefore P\left(1\right)$ is true

Let$P\left(m\right):m^3+{3m}^2+5m+3$ is divisible by 3 is true

Then, we have to prove that $p(m+1)$ is also true

$P(m+1):{(m+1)}^3+{3(m+1)}^2+5(m+1)+3$

$=m^3+1+3m(m+1)+3(m+1+2m)+5m+5+3$

$=m^3+3m^3+3m+3m^3+6m+5m+12$

$=m^3+3m^3+5m+3+3m^3+9m+9$

$=m^3+3m^3+5m+3+3(m^3+3m+3)$

$=p\left(m\right)+3(m^3+3m+3)$ which is diviesible be 3