Question

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

Answer 1

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First Factorise.

${n}^{3} - n \: = \: n( {n }^{2} - 1) \: = \: n(n + 1)(n - 1)$

6 can be expressed as 2 × 3 × 1

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n is an even number i.e, if form 2K

n^3 - n = 2k(2k+1)(2k-1)

For k belongs to Natural Numbers, it is divisible by 6

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n is an odd number ( It can be a prime number) . Taking the general case of 2K + 1

n^3 - n = (2k+1)2(k+1)2k

For k belongs to Natural Numbers, it is divisible by 6

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This can also be proved using Mathematical Induction.

Answer 2

• answer:.

replace n by 1 ,

(1)³-1= 1-1=0

0 is divisible by 6

replace n by 2

(2)³-2=8-2=6

6 is divisible by 6

replace n by 3

(3)³-3=27-3=24

24 is divisible by 6

in such way we can replace any positive integer

hence we can prove n³-n is divisible by 6 for any positive integer