Question

Prove that n x n + n is divisible by 2 for any positive integer n.​

Answer 1

Answer:

Step-by-step explanation:

when a varible is by itself it equals 1. Therefore the equation could be written...

1n x 1n + 1n

1n x 1n = 1n

1n + 1n = 2n

2n = 2                                                                                                                           2      2  

n=1

even though n=1 it's still divisible by 2

     

Answer 2

Answer:

Case i: Let n be an even positive integer.

When n=2q

In this case , we have  

n^2  −n= (2q) ^2 −2q

n^2  −n=4q^2 −2q

n^2  −n=2q(2q−1)

n^2  −n= 2r , where r=q(2q−1)

n^2  −n is divisible by 2 .

Case ii: Let n be an odd positive integer.

When n=2q+1

In this case  

n^2 −n= (2q+1)^2  −(2q+1)

n^2 −n =(2q+1)(2q+1−1)

n^2 −n =2q(2q+1)

n^2 −n =2r, where r=q(2q+1)

n^2 −n is divisible by 2.

∴ n^2 −n is divisible by 2 for every integer n