Question

plzz ans me fast...​

Answer 1

hey mate!

Step-by-step explanation:

It is true that product of two consecutive +ve integers is divisible by 2 as because of the following reasons:-

Let u consider two consecutive +ve integer as n and another as n-1.

Now, the product of both is n2-n

Case 1

When n=2q ( Since any +ve integer can be in the form of 2q or 2q+1)

n2-n= (2q)2-2q

= 4q2- 2q

It is divisible by 2 as it leaves a remainder 0 after division.

Case 2

When n= 2q+1

n2-n = (2q+1)2 - 2q+1

= 4q2+ 4q+2q+2

= 4q2+ 6q+2

It is divisible by 2

So, in both the case of the consecutive integers, it is divisible by 2

Answer 2

OK

ANY POSITIVE INTIGERS

CAN BE IN THE FORM OF

2Q,2Q+1

IF

THE FIRST NUMBER

IS 2Q THEN OTHER WILL BE 2Q+1

THERE PRODUCT

WILL BE

4Q^2+2Q=2(2Q^2+Q)

SO IT IS DIVISIBLE BY 2

NOW

IF FIRST NUMBER IS 2Q+1

THEN OTHER WILL BE 2Q+2

THEN THE PRODUCT WILL BE

4Q^2+6Q+2=2(2Q^2+3Q+1)

SO THIS ALSO DIVISIBLE BY 2

HENCE PROVED

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