# Determine if k &#x2212;<!-- − --> 1 </mrow> k </mfrac>

lilmoore11p8 2022-07-10 Answered
Determine if $\frac{k-1}{k}+\frac{1}{k\left(k+1\right)}=\frac{k}{k+1}$ holds
How to prove if the following equality holds?
$\frac{k-1}{k}+\frac{1}{k\left(k+1\right)}=\frac{k}{k+1}$
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 $k=1$
$0+\frac{1}{2}=\frac{1}{2}=\frac{1}{1+1}.$
It also holds for $k=2$
$\frac{1}{2}+\frac{1}{6}=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}=\frac{2}{2+1}$
It also works for $k=3$
$\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}=\frac{3}{3+1}$
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 $k\left(k+1\right)$ is a common multiple of $k$ and $k\left(k+1\right)$
Hence, for $k\ne 0$ and $k\ne -1$ we have
$\frac{k-1}{k}+\frac{1}{k\left(k+1\right)}=\frac{{k}^{2}-1}{k\left(k+1\right)}+\frac{1}{k\left(k+1\right)}=\frac{{k}^{2}}{k\left(k+1\right)}=\frac{k}{k+1}.$
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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.
$\frac{k-1}{k}+\frac{1}{k\left(k+1\right)}$ the difference between these two is one(on the right) is multiplied by $\left(k+1\right)$ and the other(the left) is not. To make the denominators common we multiply the numerator and the denominator of the left fraction by $\frac{k+1}{k+1}$ 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:
$\frac{k-1}{k}+\frac{1}{k\left(k+1\right)}=\frac{{k}^{2}-1}{k\left(k+1\right)}+\frac{1}{k\left(k+1\right)}$ (the 1s cancel when we add them) $=\frac{k\ast k}{k\left(k+1\right)}=\frac{k}{k+1}$ (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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