\[\frac{1}{1 \times 2} +\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+......+\frac{1}{99 \times 100}=\frac{a}{b}\] where \(\frac{a}{b}\) is in simplest fractional form. Find \(a+b\) . After finding \(a+b\), express it in the form of \(n100-x\). What is \(n+x\) ?

[Note: \(a,b,n,x\in\mathbb{Z^{+}}\) and \(0<n,x<100\)]

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