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