Fracton Twister

Let the sum be $S$ , then,

$\begin{aligned} S & = \frac{1}{1\times 2} + \frac{1}{2\times 3} + \frac{1}{3\times 4} + ... \frac{1}{99\times 100} \\ & = \sum_{k=1}^{99} \frac{1}{k(k+1)} \\ & = \sum_{k=1}^{99} \left( \frac{1}{k} - \frac{1}{k+1} \right) \\ & = \sum_{k=1}^{99} \frac{1}{k} - \sum_{k=\color{#D61F06}{1}}^{\color{#D61F06}{99}} \frac{1}{\color{#D61F06}{k+1}} \\ & = \sum_{k=1}^{99} \frac{1}{k} - \sum_{k=\color{#D61F06}{2}}^{\color{#D61F06}{100}} \frac{1}{\color{#D61F06}{k}} \\ & = \frac{1}{1} - \frac{1}{100} \\ & = \boxed{0.99} \end{aligned}$

$\frac{1}{1×2}+\frac{1} {2×3}+\frac{1}{3 \times 4}+\ldots +\frac{1}{99×100} \\ 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{99}-\frac{1}{100} \\ 1-\frac{1}{100}=\frac{99}{100}=\boxed{0.99}$

Through inspection, it can be seen that all of the terms are of the form (1/(n)(n+1)), which can be equated to (1/n)-(1/(n+1)). When all of the numbers are written in this form, all numbers except (1/1)-(1/100) get cancelled out. Solving for this gives us 99/100.