Answer: Even
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Proof:
We'll break the proof into two cases which I'll label A and B
- Case A: 'a' is even
- Case B: 'a' is odd
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Case A: 'a' is even
k = some integer
a = 2k = some even integer
a+1 = 2k+1
a(a+1) = 2k(2k+1) = 2(2k^2+k) = 2*(some integer)
Since 2 is a factor of that last expression, this shows that a(a+1) is even when 'a' is even.
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Case B: 'a' is odd
k = some integer
a = 2k+1 = some odd integer
a+1 = (2k+1)+1 = 2k+2
a(a+1) = (2k+1)(2k+2) = 2(2k+1)(k+1) = 2(some integer)
This shows that a(a+1) is even when 'a' is odd.
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Therefore, for any integer 'a', the expression a(a+1) is always even.
Some examples:
- a = 3, a+1 = 3+1 = 4, a(a+1) = 3*4 = 12 which is even
- a = 12, a+1 = 12+1 = 13, a(a+1) = 12*13 = 156 which is even
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Here's a slightly different way to interpret why the proof works.
a(a+1) consists of factors 'a' and 'a+1'
- If 'a' was even, then a(a+1) is automatically even since 2 is a factor of 'a'.
- If 'a' was odd, then a+1 is even and we arrive at the same conclusion as before.
Either way, we'll have 2 as a factor somewhere in a(a+1).