prove that the product of two consecutive natural numbers is an even umber using the principal of mathematical induction

here we are require to prove that n.(n+1) is an even number.

to prove this by mathematical induction method, we will follow three steps.

first we will show that it is true for n =1 and then we will assume that it is true for n = k

with the help of assumption we will try to prove that the given condition also holds for n = (k+1)

step 1: for n =1

n.(n+1)= 1.2 = 2 is an even number

step 2: let us assume that k(k+1) is an even number. i.e. either k or (k+1) is an even number.

step 3: for (k+1)(k+2) , by the assumption two cases arise:

case I: if k is an even number, i.e. k =2m , where m is an integer

then k+2=2m+2=2(m+1) is also an even number, therefore (k+1).(k+2) is also an even number

case II: if (k+1) is an even number. (k+1).(k+2) is an even number, as the multiplication of an even number with any other integer result into an even integer.

thus (k+1)(k+2) is an even integer. thus it is true for any integer n.

thus the product of two consecutive natural number is an even integer.

hope this helps you

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