Question

For any positive integer
n, prove that n3−n is
divisible by 6.

Answer 1

HOLA !!

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The integer N can be written as any number to convenience

To prove :- n3 - N ÷ 6 is a positive integer

Let's assume N to be 1 , 2 ,3

First case

n3 - N = 3 ( 3) - 3 = 6

3 (3) - 3 is divisible by 6

Second case

N3 - N = 1 ( 3) - 1 = 2

2 is divisible by 6

Third case

N3 - N = 2 ( 3) - 4 = 2

2 is divisible by 6

Hence , For any integer n3 - N is divisible 6

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HOPE U UNDERSTAND

Answer 2

HELLO DEAR ,

HERE IS YOUR ANSWER,

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To Prove :- For any positive integer n, n^3-n is divisible by 6.


Proof : Let n be any positive integer.

$= > {n}^{3} - n = (n - 1)(n)(n + 1)$

Since any positive integer is of the form 6q or 6q+1 or 6q+2 or 6q+3 or 6q+4 , 6q+5


●If n = 6q

Then , (n-1) n (n+1) = (6q-1) 6q (6q+1)

=> which is divisible by 6.


●If n = 6q+1

Then, (n-1) n (n+1) = (6q) (6q+1) (6q+2)

=> which is divisible by 6.


● If n = 6q+2

Then, (n-1) n (n+1) = (6q+1) (6q+2) (6q+3)

=> (n-1) n (n+1) = 6 (6q+1) (3q+1) (2q+1)


=> which is divisible by 6


Similarly, we can prove others.

●Hence it is proved that for any positive integer n, n^3-n is divisible by 6.


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Hope it helps u dear !!!!!

# Nikky ✌ ✌