If r is an integer greater than 1, what is the value of (−1)^r +1 if:
1. r is an odd integer
2.r is an even integer
(-1)^odd integer = -1
(-1)^even integer = +1
so you have either
-1 + 1 =
or
1 + 1=
You can use trial and error to work this out.
Just sub in points
1. choose r to be odd and see what you obtain : )
Ah, the wonderful world of integers! Let's tackle these questions with a touch of humor.
1. If r is an odd integer, well, -1 raised to any odd power is always -1, my friend. So, when we add 1 to it, we get 0. Why? Because -1 is feeling a bit down, and when we add 1 to it, it perks right up, making everything balanced in the realm of mathematics.
2. Now, when r is an even integer, things get a little more interesting. -1 raised to any even power is always 1. It's like a magician, doing the same trick over and over again, always ending up with the same result. So when we add 1 to it, we get 2. Magic!
In conclusion, the value of (-1)^r + 1 is 0 when r is odd and 2 when r is even. So, depending on whether r likes odd or even numbers, we have our answers!
To find the value of (−1)^r + 1 for different values of r, we can apply the properties of exponents and simplification rules for negative numbers:
1. If r is an odd integer:
When r is an odd integer, it can be represented as r = 2k + 1, where k is an integer.
Substituting this value in (−1)^r + 1:
(−1)^(2k + 1) + 1
We know that (−1)^3 = −1, (−1)^5 = −1, (−1)^7 = −1, and so on.
Therefore, applying this pattern:
(−1)^r + 1 = −1 + 1 = 0.
So, when r is an odd integer, the value of (−1)^r + 1 is 0.
2. If r is an even integer:
When r is an even integer, it can be represented as r = 2k, where k is an integer.
Substituting this value in (−1)^r + 1:
(−1)^(2k) + 1
We know that (−1)^2 = 1, (−1)^4 = 1, (−1)^6 = 1, and so on.
Therefore, applying this pattern:
(−1)^r + 1 = 1 + 1 = 2.
So, when r is an even integer, the value of (−1)^r + 1 is 2.