Show that \( 2.5^{n}-5^{n+1}+5^{n+2} \) is even for all positive integer values of \( n \)

To show that \( 2.5^{n}-5^{n+1}+5^{n+2} \) is even for all positive integer values of \( n \), we start by simplifying the expression: \[ 2.5^{n} - 5^{n+1} + 5^{n+2} = 2.5^{n} - 5 \cdot 5^{n} + 25 \cdot 5^{n} \] \[ = 2.5^{n} - 5^{n+1} + 5^{n+2} = 2.5^{n} + 20 \cdot 5^{n} \] \[ = 2.5^{n} + 5^{n+1} \cdot 4 \] Now, we rewrite using properties of powers of 5: The expression can be rewritten as follows: \[ = 2.5^{n} + 4.5^{n+1} \] Next, we factor out \( 5^{n} \): \[ = 5^{n}(2 + 4 \cdot 5) \] Now notice that \( 5^{n}(2 + 20) = 5^{n} \cdot 22 \). Since \( 5^{n} \) is odd for any integer \( n \) and \( 22 \) is even, the product of an odd number and an even number is always even. Therefore, \( 2.5^{n}-5^{n+1}+5^{n+2} \) is even for all positive integer values of \( n \). Thus, we conclude: \[ \boxed{\text{Even for all } n \geq 1.} \]