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lilmoore11p8 2022-07-10 Answered
Determine if k 1 k + 1 k ( k + 1 ) = k k + 1 holds
How to prove if the following equality holds?
k 1 k + 1 k ( k + 1 ) = k k + 1
Maybe finding a common denominator would work, but I have no idea how to do it in this example.
I see that it holds for k = 1
0 + 1 2 = 1 2 = 1 1 + 1 .
It also holds for k = 2
1 2 + 1 6 = 3 6 + 1 6 = 4 6 = 2 3 = 2 2 + 1
It also works for k = 3
2 3 + 1 12 = 8 12 + 1 12 = 9 12 = 3 4 = 3 3 + 1
But I am not sure how to proceed if I am not working with numbers but with expressions.
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Answers (2)

postojahob
Answered 2022-07-11 Author has 13 answers
Notice that k ( k + 1 ) is a common multiple of k and k ( k + 1 )
Hence, for k 0 and k 1 we have
k 1 k + 1 k ( k + 1 ) = k 2 1 k ( k + 1 ) + 1 k ( k + 1 ) = k 2 k ( k + 1 ) = k k + 1 .
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Yesenia Obrien
Answered 2022-07-12 Author has 5 answers
You're having trouble with finding the common denominator. Let's compare the two denominators.
k 1 k + 1 k ( k + 1 ) the difference between these two is one(on the right) is multiplied by ( k + 1 ) and the other(the left) is not. To make the denominators common we multiply the numerator and the denominator of the left fraction by k + 1 k + 1 which if you think about it, is the same as multiplying it by one, which does not affect the value of the fraction. Doing this we get:
k 1 k + 1 k ( k + 1 ) = k 2 1 k ( k + 1 ) + 1 k ( k + 1 ) (the 1s cancel when we add them) = k k k ( k + 1 ) = k k + 1 (We cancel the ks on the top and bottom)
It is important that this equality holds almost everywhere. We cannot say it is valid at k = 0 , 1 this is because these values of k result in division by 0 which we know is not allowed.
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