$\frac{1}{1 \times 2} +\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+......+\frac{1}{99 \times 100}=\frac{a}{b}$ 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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