@the-blade-dancer The second step looks quite different from the first step! It's because for the first two fractions, we have multiplied the top and bottom by a factor.
For the first fraction, we have multiplied the top and bottom by \( n + 1.\)
\(\frac{1}{\textcolor{red}{n}} = \frac{n + 1}{\textcolor{red}{n} (n + 1) } \)
For the second fraction, we have multiplied the top and bottom by \(n.\)
\(\frac{1}{\textcolor{red}{n+1}} = \frac{n}{\textcolor{red}{(n +1)} n } \)
For the third fraction, we've done something different: we have factored the bottom. (Factor just means to write as a product of two terms, like \( 12 = 3 \times 4.\) )
\( - \frac{1}{n^2 + n} = - \frac{1}{n(n+1)} \)
Why are we trying to do this? To make the bottoms of the fractions all equal to \( n(n+1).\) Then we can add the fractions together very easily, and luckily we have some terms cancel out, so we get \(0\) on the top, which is a little fortunate!
\( \frac{n + 1 - n - 1}{n(n+1)} \)
\( = \frac{0}{n(n+1)} = \boxed{0}.\)
🙂
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