Question

For any positive integer n ,prove that n^3-n is divisible by 6

Answer 1

Answer:

$let \: f(n) = {n}^{3} - n$

$now \: f(1) = 0 \: divisible \: by \: 6$

$again \: f(2) = 6 \: divisible \: by \: 6.$

So the given statement is true for n=1, 2.

Let us take the statement is true for some integer k.

$f(k) = {k}^{3} - k \: is \: divisible \: by \: 6$

$let \: f(k) = 6k_{1} \ \: ( k _{1} \: being \: an \: integers)....(1)$

Now,

$f(k + 1)$

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

$= {k}^{3} + {3k}^{2} + 2k$

$( {k}^{3} - k) + 3k(k + 1).....(2)$

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

k(k+1) is the product of two consecutive integers, which is always divisible by 2.

Then we can write k(k+1)=2k2 [Where k2is some integer.].

Using this and (1) form (2) we get,

$f(k + 1) = 6k _{1} + 6k _{2} = 6(k _{1} \: + \: k _{2})$

k1,k2 are integers then k1+k2 is also an integer.

So f(k+1) is also divisible by 6.

So the statement is true for n=k+1 if we assume it to be true for n=k.

By the principle of mathematical induction the statement is true ∀n∈N.