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## Answers

ii) Then their product is P = n(n+1)

iii) For n = 1, P = 1*2 = 2, which is divisible by 2; so it is true for n = 1

Thus the statement is k(k+1) is divisible by 2 for k a positive integer ----------- (1)

P(k+1): (k+1)(k+1+1) = (k+1)(k+2) = k(k+1) + 2(k+1)

==> P(k+1) = P(k) + 2*(A positive integer) [From (1), P(k) = k(k+1)]

Thus both terms are divisible by 2

Hence P(k+1) is also divisible by 2 --------------- (2)

Thus from (1) & (2), for two consecutive arbitrary terms, it is proved that the product is divisible by 2. Hence the product of two consecutive positive integers is divisible by 2.