# Simple algebra development problem.

I have a simple development of an expression that I can not understand how it was made. The expression is:

\begin{align}\frac{1}{k(k+1)} & = \frac{1}{k} - \frac{1}{k+1} \end{align}

Can please someones show me how the left side is equal to the right side?

Thank you!

• First introduction to partial fractions? – Mike Oct 24 '12 at 14:31
• This is an example of a partial fraction expansion of a rational function, which is easily calculated by the Heaviside cover-up method. It deserves to be better known that this method also works for nonlinear denominators. – Bill Dubuque Oct 24 '12 at 16:29
• Thanks for you answers! Partial fractions are actually introduced later in my book. Still it was good to read up on them! – Lukas Arvidsson Oct 25 '12 at 8:05

When adding two fractions, it is useful to use common denominator.

In your case, $k(k+1)$ is a multiple of both $k$ and $k+1$, so you can write:

$$\frac1k-\frac1{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k(k+1)}.$$

A slightly different approach: If you are given $\frac1{k(k+1)}$ and you want to simplify it, you may notice that $\frac{k+1}{k(k+1)}=\frac1k$ and $\frac{k}{k(k+1)}=\frac1{k+1}$ are simpler expressions. So you can ask whether you can somehow write the numerator using $k+1$ and $k$. And you can: $1=(k+1)-k$.

So you get $$\frac1{k(k+1)} = \frac{(k+1)-k}{k(k+1)}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)}=\frac1k-\frac1{k+1}.$$

• Thank you! This was probably the way I was supposed to do it according to my book. Many thanks! – Lukas Arvidsson Oct 25 '12 at 8:04

Recall how fractions are subtracted: $$\frac{a}{b} - \frac{c}{d} = \frac{a \cdot d}{b \cdot d} - \frac{b \cdot c}{b \cdot d} = \frac{a \cdot d- b \cdot c}{b \cdot d}$$ In the case at hand $a=c=1$, $b=k+1$ and $d=k$.

• Thank you for your answer! – Lukas Arvidsson Oct 25 '12 at 8:06