Question

Prove by the Principle of Mathematical Induction: n^3−7n + 3 is divisible by 3, for all natural numbers n.

Answer 1

Assume given statement

Let $P\left(n\right)=n^3-7n+3$ is divisible by 3, for all natural numbers n.

Check that statement is true for $n=1$

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

$\Rightarrow P\left(1\right)$is divisible by 3

So, P(n) is true for$n=1$.

Assume P(k) to be true and then prove $P(k+1)$ is true.

Assume that P(k) is true for some $n=kϵN$

So, $P\left(k\right):k^3-7k+3$ is divisible by 3

$\Rightarrow k^3-7k+3=3q,$


where $qϵN\dots \left(1\right)$

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

$=k^3+1+3k^2+3k-7k-7+3$

$=k^3-7k+3+3k^2+3k-6$

$=k^3-7k+3+3(k^2+k-2)P(k+1)$

$=3q+3(k^2+k-2)$ (From (1))

$=3[q+k^2+k-2]$

So, $P(k+1)$ is divisible by 3.

Thus, P(k+1) true whenever P(k) is true.

Hence, By Principle of mathematical Induction, P(n) is true for all natural numbers n.