Question

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

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Answer 1

Answer:

Let n be an arbitrary natural number

On dividing n by 6 we get q as quotient and r as remainder. Where 0≤r≥6

n=6q+r

Case 1

when r=0

n=6q

n^3-n=(6q)^3-6q

6(36q^3-q)

It is divisible by 6

Similarly when r=1,2,3,4,5 then n^3-n is divisible by 6.

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Answer 2

Answer:

Let f(n) = n³-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) = k³ − k is divisible by 6.

Let f(k) = 6k1 [k1 being an integer].......(1).

Now

f(k + 1)

=(k + 1)³ − (k + 1)

=k³ + 3k² + 2k

=k³ − k + 3k² = 3k

=(k³−k) + 3k(k + 1).........(2).

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 k2 is some integer.].

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

f(k + 1) = 6k1 + 6k2 = 6(k1 + k2)

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 when ∀n ∈ N.

$\huge{ \bold{ \underline{ \green{ \sf{Hence \: Proved}}}}}$