Question

Prove that n³ $-$ 7n + 3 is divisible by 3 for all n $\in$ N.

Answer 1

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

Hence, n³ $-$ 7n + 3 is divisible by 3 for all n $\in$ N.