Question

prove that the product of two consecutive positive integer is divisible by 2​

Answer 1

Answer :-

We have to prove that the product of two consecutive integers is divisible by 2 .

Let the integer be m

Then another integer = m +1

Now by multiplying

= m(m + 1)

= m² + m

Now if m is even it is in the form of 2n

= (2n)² + 2n

= 4n² + 2n

= 2n( 2n + 1)

Which is always divisible by 2

Now if m is odd it is in the form of 2n + 1

= (2n + 1)² + (2n + 1)

= (4n² + 2n + 1) + (2n + 1)

= 4n² + 4n + 2

= 2( 2n² + 2n + 1)

Which is always divisible by 2

Hence Proved

Answer 2

Answer:

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Let n and n + 1 are two consecutive positive integer

We know that n is of the form n = 2q and n + 1 = 2q + 1

n (n + 1) = 2q (2q + 1) = 2 (2q2 + q)

Which is divisible by 2

If n = 2q + 1, then

n (n + 1) = (2q + 1) (2q + 2)

= (2q + 1) x 2(q + 1)

= 2(2q + 1)(q + 1)

Which is also divisible by 2

Hence the product of two consecutive positive integers is divisible by 2