Question 984151: I would appreciate your help!
If k stands for an integer, then is it possible for k^2+k to stand for an odd integer? Explain.
Found 3 solutions by josgarithmetic, reviewermath, Alan3354:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! Not Possible.
Both terms are odd if k is odd;
Both terms are even if k is even.
Sum of even integers is even.
Sum of TWO odd numbers is even.
Answer by reviewermath(1029)  (Show Source):
You can put this solution on YOUR website! If k stands for an integer, then is it possible for k^2+k to stand for an odd integer? Explain.
The answer is "NO".
Case 1: k is an odd integer
For some integer p, k = 2p + 1
So, k + 1 = 2p + 2 = 2(p+1) = 2q for some integer q
k^2 + k = k(k+1) = k(2q) = 2kq = 2r for some integer r.[Integers are closed under multiplication]
Therefore, if k is an odd integer, then k^2 + k is an even integer
case 2: k is an even integer
For some integer p, k = 2p
So, k + 1 = 2p + 1
k^2 + k = k(k+1) = 2p(2p+1) , multiple of 2
Therefore, if k is an even integer, then k^2 + k is an even integer
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! If k stands for an integer, then is it possible for k^2+k to stand for an odd integer? Explain.
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No, not possible.
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k^2 + k = k*(k+1)
Either k or k+1 is even --> the product is even.
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