How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mess with the constant edits, I will stop editing and not touch this so people can review the question. Sorry
a) Prove that P(1) is true
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$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$
Show that P(k+1) is true as well
$$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$
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$$ = \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$
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$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
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$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
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$$ = \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$
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$$ = \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$

Hint:
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{(k+1)(k+1)}{(k+1)(k+2)}.$$

Regarding your calculations, note that
$$ \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)} = \frac{k^2+2k+1-k^2-2k}{(k+1)(k+1+1)}=?$$
