Question

Given that n is an integer and that n>1, prove algebraically that n²-(n-2)²-2 is always an even number

Answer 1

Answer:

see explanation

Step-by-step explanation:

Any integer n > 1 multiplied by 2 will be even, that is

2n ← is even

Given

n² - (n - 2)² - 2 ← expand parenthesis

= n² - (n² - 4n + 4) - 2

= n² - n² + 4n - 4 - 2 ← collect like terms

= 4n - 6 ← factor out 2 from each term

= 2(2n - 3)

Hence 2(2n - 3) ← will always be even for n > 1

Answer 2

The statement can be proved by parenthesis that the expression $n^{2} - (n - 2)^{2} - 2$ is always an even number.

What is the process of expanding the polynomial and check whether it is a even number ?

=  $n^{2} - (n - 2)^{2} - 2$

=  $n^{2} - (n^{2} - 4n + 4) - 2$

=  $4n - 4 - 2$

=  $4n - 6$

= $2(2n - 3)$

As the expression 2(2n - 3) is always a multiple of 2 and also it is given that n>1 therefore by the parenthesis, the expression $n^{2} - (n - 2)^{2} - 2$ is always an even number.

To learn more about algebraic expressions, refer -

https://brainly.com/question/2164351

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