Question

prove that n²-n is divisible by 2.​

Answer 1

Here is the answer to your question.

 Any positive integer is of the form 2q or 2q + 1, where q is some integer

When n = 2q, 

n²+n=(2q)²+2q

      =4q²+2q

      = 2q(2q+1)   

which is divisible by 2

 

when n=2q+1

n²-n=(2q+1)²+(2q+1)

      = 4q²+4q+1+2q+1

      =4q²+6q +2

     =  2(2q²+3q+1)

which is divisible by 2

hence n²+n is divisible by 2 for every positive integer n

Hop it will help u and don't forget to mark as brainlist....

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Answer 2

• Let positive integers be n and n + 1

We have to prove that n² - n is divisible by 2.

Here .. n = 2

a = bq + r or nq + r

When n = 2q

=> n² - n = (2q)² - 2q

=> 4q² - 2q

=> 2q(2q - 1)

Divisible by 2

When n = 2q + 1

=> n² - n = (2q + 1)² - 2q + 1

=> 4q² + 1 + 4q - 2q + 1

=> 4q² + 2q + 2

=> 2(2q² + 2q + 1)

Divisible by 2

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Hence, for any position integer n .. n² - n is divisible by 2.

______ [HENCE PROVED]

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