Question

Prove by the Principle of Mathematical Induction: $n^3-n$ is divisible by 6, for each natural number $n\ge 2.$

Answer 1

Assume given statement

Let $P\left(n\right)=n^3-n$ is divisible by 6,

for each natural number $n\ge 2$.

Check that statement is true for

$n=2P\left(2\right)=2^3-2=8-2=6$

$\Rightarrow P\left(2\right)$ is divisible by 6

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

Lets assume P(k) is true

So, P(k) is divisible by 6

$\Rightarrow k^3-k=6q,$ where $qϵN\dots \left(1\right)$

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

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

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

$=k^3-k+3k^2+3kP(k+1)$

$=6q+3k(k+1)$

Now if k is odd $k+1$ will be even and if k is even $k+1$ will be odd and even multiplied with odd is always even, so $k(k+1)=2p$

$\Rightarrow P(k+1)=6q+3.2p=6q+6p=6(p+q)$

$\therefore P(k+1)$ is divisible by 6

Hence $P(k+1)$ is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers $n\ge 2.$