Question

show that n(n+1)(n-1) is divisible by 3

Answer 1

Let P(n) : n(n+1)(n-1) is divisible by 3.

P(1) = 1 × 2 × 0 = 0 is divisble by 3

Hence P(1) is true.

P(k) = k (k+1) (k-1)

= k(k-1)(k-2) + 3[k(k-1)]

Here, term 3[k(k-1)] is divisible by 3.

And term k(k-1)(k-2) is divisible by 6 and 6 is divisble by 3.

Hence k(k-1)(k-2) + 3[k(k-1)] is divisible by 3

P(k) is true.

REASON FOR k(k-1)(k-2) DIVISIBLE BY 6 :

▪k(k-1)(k-2) is three consecutive integers.

▪Every alternate number is even and divisible by 2 and Every alternate third number is divisible by 3.

▪Hence, If you are to choose any 3 consecutive integers, it's product will be divisible by 3.

For every P(n) we assume P(k) is true. So the statement is true.

So n(n+1)(n-1) is divisible by 3.