Question

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

Answer 1

SOLUTION :

Let n be any positive integer. n³ -  n = (n - 1)   (n) (n + I)

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

Case : 1

If n = 6q,  then

(n -1) n (n + 1)  = (6q -1)  6q (6q + 1),Which is divisible by 6

Case : 2

If n = 6q + 1,  then,

(n -1) n (n + 1) = (6q) (6q + 1) (6q + 2),  Which is divisible by 6.

Case : 3

If n = 6q + 2,  then,

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

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

(n -1) n (n + 1) = 6 (6q + 1) (3q + 1) (2q + 1),Which is divisible by 6.

Similarly, n (n + 1) (n + 2) is divisible by 6 if n=  6q + 3 or 6q + 4, 6q + 5.

Hence it is proved that for any positive integer n, n³ - n is divisible by 6.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

hello mate

here is your solution

 Given that S(n) = n3-n divisible by 6.

Let n =1 then we get '0' which is divisible by 6.  
∴ S(1) is true. Let us assume that
n = k

now
  S(k) = k3- k which is divisible by 6.
 ∴ (k3-k) / 6 = m ( integer )  
(k3-k)  = 6m   
k3= 6m +k --------→(1)

now we have to prove that n = k+1 
⇒ (k+1)3 - (k+1
) ⇒ (k3+3k2+3k+1) - (k+1)
now
subsitute from equation (1) in above equation then
 ⇒ 6m +k+3k2+2k 
⇒ 6m +3k2+k 
⇒ 6m +3k(k+1)  ( ∴k(k+1) = 2p is an even number p is natural number)

 ⇒ 6m +3x2p
 ⇒ 6(m +p
∴which is divisible by 6 s(k+1) is correct.

I hope helps you