Part (1)
n is some positive integer. Let's say for now that n is even. So n = 2k, for some integer k
This means n-1 = 2k-1 is odd since subtracting 1 from an even number leads to an odd number.
Now multiply n with n-1 to get
n(n-1) = 2k(2k-1) = 2m
where m = k(2k-1) is an integer
The result 2m is even showing that n(n-1) is even
------------
Let's say that n is odd this time. That means n = 2k+1 for some integer k
And also n-1 = 2k+1-1 = 2k showing n-1 is even
Now multiply n and n-1
n(n-1) = (2k+1)(2k) = 2k(2k+1) = 2m
where m = k(2k+1) is an integer
We've shown that n(n-1) is even here as well.
------------
So overall, n(n-1) is even regardless if n is even or if n is odd.
Either n or n-1 will be even. If you multiply an even number with any number, the result will be even.
=======================================================
Part (2)
n is some positive integer
2n is always even since 2 is a factor of 2n
2n+1 is always odd because we're adding 1 to an even number. The sequence of integers goes even,odd,even,odd, etc and it does this forever.
-----------
Another way to see how 2n+1 is odd is to divide 2n+1 over 2 and you'll find that we get (2n+1)/2 = 2n/2+1/2 = n+0.5
The 0.5 at the end is not an integer, so there's no way that (2n+1)/2 is an integer; therefore 2n+1 is odd.
, because factoring after all is a form of division. When we divide a fraction by a fraction, we bring the lower fraction up and flip it to multiply, giving us this: 
. This of course reduces to 4. So when we factor out 1/8, what we will have then is 
, choice D from above.
