<s>[INST] Question on algebra used in induction proof Can someone take a look at this simplification... it is part of a proof by induction solution I'm examining... I'm interested in the result of this simplification as I'm not sure how the third to last step was achieved.
$$\begin{align*}
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}&=\\
\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{k(k+1)}\right)+\frac{1}{(k+1)(k+2)}&=\\
\left(1-\frac{1}{k+1}\right)+\frac{1}{(k+1)(k+2)}&=\\
1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}&=\\
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}&=\\
1-\frac{k+1}{(k+1)(k+2)}&=\\
1-\frac{1}{k+2}&=\\
1-\frac{1}{(k+1)+1}.
\end{align*}$$
Now to me it seems that the third to last step would result in
$$\begin{align*}
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}&=\\
\bbox[yellow]{
1-\frac{k+3}{(k+1)(k+2)}
}&=
\end{align*}$$
What am I missing? 
 [/INST] That step should be
$$
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
=1+(-1)\cdot\left(\frac{k+2}{(k+1)(k+2)}\right)+\frac{1}{(k+1)(k+2)}\\
=1+\frac{-k-2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
=1+\frac{-k-1}{(k+1)(k+2)}\\
=1-\frac{k+1}{(k+1)(k+2)} \\
=1-\frac{1}{k+2}.
$$
</s>