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$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots \ldots \ldots+\ldots \ldots \cdot \frac{1}{n \cdot(n+1)}$
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Verified Answer
The correct answer is:
$\frac{n}{n+1}$
$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots \ldots \ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$
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