Math, asked by anu5hka12, 1 year ago

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

Answers

Answered by Anonymous
49

Step-by-step explanation:


n³ - n = n (n² - 1) = n (n - 1) (n + 1)



Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.



∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.


If n = 3p, then n is divisible by 3.


If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.



If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.



So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.


⇒ n (n – 1) (n + 1) is divisible by 3.


Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.



∴ n = 2q or 2q + 1, where q is some integer.



If n = 2q, then n is divisible by 2.


If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.


So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.



⇒ n (n – 1) (n + 1) is divisible by 2.


Since, n (n – 1) (n + 1) is divisible by 2 and 3.



∴ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)




THANKS


#BeBrainly.


Answered by fanbruhh
20
 \huge \bf{ \red{hey}}

 \huge{ \mathfrak{ \blue{here \: is \: answer}}}

For any positive integer n prove that n cube minus n is divisible by 6

when any number is divided by 3 the remainder is 0, 1, 2

now

according to question

hence, n = 3p or
3p + 1 or
3p + 2,

here p is some integer.

n³ - n = n (n² - 1) = n (n - 1) (n + 1)

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

so we can say

one of any positive integer

n, n-1, n+1 is divisible by 3

⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n is divisible by 2.

If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

⇒ n (n – 1) (n + 1) is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

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 \huge{ \mathfrak{ \pink{thanks}}}
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