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$\frac{3(\sum\limits_{n=4}^\infty \frac{(\lim_{x \to +\infty} \frac{1+2+3+...+x}{x ^ 2})}{2 ^ n})}{5}=\frac{a}{b}$

$gcd(a, b)=1$

$ab-1=x$

What is the value of $x$?

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