If n is a positive integer and n > 1, prove that (n over 2)+ (n-1 over 2) is a perfect square.
To prove that (n over 2) + ((n-1) over 2) is a perfect square, we can start by simplifying the expression.
Let's first consider (n over 2). The expression (n over 2) represents the number of ways to choose 2 items from a set of n items. This can be calculated using the formula:
(n over 2) = n * (n-1) / 2
Similarly, let's consider ((n-1) over 2). This expression represents the number of ways to choose 2 items from a set of n-1 items.
((n-1) over 2) = (n-1) * (n-2) / 2
Now, let's simplify the expression (n over 2) + ((n-1) over 2) by substituting the above formulas:
(n over 2) + ((n-1) over 2) = (n * (n-1) / 2) + ((n-1) * (n-2) / 2)
To further simplify the expression, let's find a common denominator and combine the terms:
(n * (n-1) / 2) + ((n-1) * (n-2) / 2) = [(n * (n-1)) + ((n-1) * (n-2))] / 2
Expanding and combining like terms:
[(n * (n-1)) + ((n-1) * (n-2))] / 2 = (n^2 - n + n - n^2 + 2n - 2) / 2
Simplifying:
(n^2 - n + n - n^2 + 2n - 2) / 2 = (2n - 2) / 2 = n - 1
Now we have simplified the expression to n - 1.
To prove that n - 1 is a perfect square, we can set it equal to a perfect square and solve for n:
n - 1 = m^2 (let's assume n - 1 is a perfect square, represented by m^2)
Rearranging the equation:
n = m^2 + 1
Since n is a positive integer, we need to show that m^2 + 1 is always a perfect square. One way to do this is by considering the difference of consecutive perfect squares:
(m+1)^2 - m^2 = (m^2 + 2m + 1) - m^2 = 2m + 1
We observe that the difference of consecutive perfect squares is always 2m + 1. Therefore, for any positive integer m, 2m + 1 is always an odd number.
Since m is an integer, (2m + 1) is odd, and adding an odd number to an even number always results in an odd number. Thus, m^2 + 1 is always an odd number.
Since n = m^2 + 1, n must also be an odd number.
Now, let's substitute the value of n into the expression n - 1:
n - 1 = (m^2 + 1) - 1 = m^2
We see that n - 1 is equal to m^2, which is a perfect square.
Therefore, we have proven that (n over 2) + ((n-1) over 2) is indeed a perfect square, as it simplifies to n - 1 which is a perfect square.