Solve the following
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 $n \in \mathbf{N}$.