Question

$n^3-7n+3$ is divisible by 3 for all $nϵN$.

Answer 1

Let P(n) : $n^3-7n+3$ is divisible by 3, for all natural number n.

We observe that P(1) is true.

$P\left(1\right)=(1{)}^3-7(1)+3$

$=1-7+3=-3$, Which is divisible by 3,

Hence, P(1) is true.

Now, assume that P(n) is true for n = k.

$P\left(k\right)=k^3-7k+3=3q$

Now, We shall prove P(k + 1) is true

$P(k+1):(k+1{)}^3-7(k+1)+3$

$=k^3+1+3k(k+1)-6$

$3q+3\left[k\right(k+1)-2]$

Hence, P(k + 1) is true whennever P(k) is true.

So, By the principle of mathematical induction

P(n) : is true for all natural numbers n.