Explain why n(n-1) must be an even number​

n & n - 1 are two consecutive numbers which means one of is surely even and the other one is odd.

Suppose that n is an even number;

So we can right n like this :

n = 2t

Thus :

n - 1 = 2t - 1

n( n - 1 ) = 2t( 2t - 1 ) = 2 ( 2t^2 - t )

= 2 ( q ) = 2q

The number obtained is a multiple of 2, so it is definitely even because all numbers are even multiples of 2.

answer: the common difference is -0.10