Question

pls say.....................

Answer 1

Heya !!!
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
 S(k) = k3- k
which is divisible by 6.
∴ S(k) is true.
∴ (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)
subsitute 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 true.

Answer 2

n^3-n = n(n^2-1) = n*(n-1)*(n+1) or, (n-1)*n*(n+1).
This means, its the product of 3 consecutive numbers.
The product of 3 consecutive numbers is always divisible by 6.
For eg: 1*2*3 or 7*8*9 or 101*102*103 etc