Question

for any positive integer n,prove that n3_n divisible by 6​

Answer 1

Step-by-step explanation:

LET THE N BE 1

THEN,

N^3–N=0

0÷6=0

N=2

N^3–N=6

6÷6=1

N=3

N^3–N=24

24÷6=4

SO,

PROOVED THAT FOR ANY POSITIVE INTEGER N, THEN N^3–N IS DIVIDED BY 6

PROOVED

PLEASE MARK AS A BRAINLIST

Answer 2

Answer:

Step-by-step explanation:

$n^{3}-n$

when we factoize the same

we get

$n(n^{2}-1)$

we know that $x^{2} - y^{2} = (x+y)(x-y)$   - - - - - - (algebraic identity)

therefore we get

(n-1)(n)(n+1)

now, we will observe that if

Case 1) n is odd, then n-1 or n+1 will be even and divisible by 2

Case 2) n is even, then the n in the expression is also even and divisible by 2

therefore the expression will always be divisible by 2.

now, if n is of the form

Case 1) 3n : Then, the n in the expression is also of the form 3, hence divisible by 3

Case 2) 3n+1 : Then, n-1 will be of the form 3n+1-1 = 3n, making the expression divisible by 3

Case 3) 3n+2 : Then n+1 will be of the form 3n+2+1= 3n+3 = 3(n+1) again making the expression divisible by 3.

Now, we know that the expression is always divisible by 2 and 3

THEREFORE, it will always be divisible by 6 (as both are co prime to each other)

CHEERS MATE, PLEASE MARK BRAINLIEST