Question:
Prove that $n^{3}-7 n+3$ is divisible by 3 for all $n \in \mathbf{N}$.
Solution:
Let $\mathrm{p}(n)=n^{3}-7 n+3$ is divisible by $3 \forall n \in \mathbf{N}$.
Step I : For $n=1$,
$\mathrm{p}(1)=1^{3}-7 \times 1+3=1-7+3=-3$, which is clearly divisible by 3
So, it is true for $n=1$
Step II : For $n=k$,
Let $\mathrm{p}(k)=k^{3}-7 k+3=3 m$, where $m$ is any integer, be true $\forall k \in \mathbf{N}$.
Step III : For $n=k+1$
$\mathrm{p}(k+1)=(k+1)^{3}-7(k+1)+3$
$=k^{3}+3 k^{2}+3 k+1-7 k-7+3$
$=k^{3}+3 k^{2}-4 k-3$
$=k^{3}-7 k+3+3 k^{2}+3 k-6$
$=3 m+3\left(k^{2}+k+2\right) \quad[$ Using step II]
$=3\left(m+k^{2}+k+2\right)$
$=3 p$, where $p$ is any integer
So, $\mathrm{p}(k+1)$ is divisible by 3 .
Hence, $n^{3}-7 n+3$ is divisible by 3 for all $n \in \mathbf{N}$.