Proof by induction - fractions

mikky05v
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Homework Statement


I have been working on this proof for a few hours and I can not make it work out.

$$\sum_{i=1}^{n}\frac{1}{i(i+1)}=1-\frac{1}{(n+1)}$$

i need to get to
$$1-\frac{1}{k+2}$$

I get as far as
$$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
then I have tried
$$1-\frac{(k+2)+1}{(k+1)(k+2)}$$
by multiplying the left fraction by (k+2) which got me nowhere.

What am I doing wrong?
 
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disregard, I figured it out. I simply had to separate 1/i(i+1) into 1/i - 1/(i+1)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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