Question

prove that product of two consecutive positive integers is divisible by 2

Answer 1

let n and n-1 be the 2 positive integers.product= n(n-1)=n2-n
CASE 1 (when n is even)Let n=2qn2-n=(2q)2-2q=4q2-2q=2q(2q-1)Hence the product n2-n is divisible by 2
CASE 2(when n is odd)Let n be 2q+1n2-n= (2q+1)2- (2q+1)=4q2+4q+1-2q-1=4q2+2q=2q(2q+1)Hence the product n2-n is divisible by 2

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