We can prove by induction that the expression 2.5^n - 5^(n+1) + 5^(n+2) is always even for all positive integer values of n.
Base case:
For n = 1, the expression becomes:
2.5^1 - 5^2 + 5^3 = 2.5 - 25 + 125 = 102.5
Since 102.5 is not an integer, but a decimal number, the base case does not hold true.
Inductive step:
Assume that for some positive integer k, the expression 2.5^k - 5^(k+1) + 5^(k+2) is an even number.
We need to prove that for k+1, the expression 2.5^(k+1) - 5^(k+2) + 5^(k+3) is also even.
Starting with the defined expression for k:
2.5^k - 5^(k+1) + 5^(k+2)
To find the expression for k+1, multiply each term by 2.5:
2.5*(2.5^k) - 2.5*(5^(k+1)) + 2.5*(5^(k+2))
This simplifies to:
2.5^(k+1) - 5^(k+2) + 5^(k+3)
Now, it is clear that the expression for k+1 has the form of the expression for k, just multiplied by 2.5. Since the expression for k is known to be even, 2.5*(even number) will still result in an even number.
Therefore, by induction, the expression 2.5^n - 5^(n+1) + 5^(n+2) is always even for all positive integer values of n.
show that 2.5^n - 5^n+1 + 5^n+2 is even for all positive integer value
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