The value of 1000 left[ frac{1}{1 times 2} + | vedclass

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(B) The given expression is $1000 \left[ \sum_{n=1}^{999} \frac{1}{n(n+1)} \right]$.Using the method of partial fractions, we know that $\frac{1}{n(n+

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$999$

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(B) The given expression is $1000 \left[ \sum_{n=1}^{999} \frac{1}{n(n+1)} \right]$.Using the method of partial fractions, we know that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.Substituting this into the sum, we get $1000 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \ldots + (\frac{1}{999} - \frac{1}{1000}) \right]$.This is a telescoping series where all intermediate terms cancel out, leaving $1000 \left[ 1 - \frac{1}{1000} \right]$.$= 1000 \left[ \frac{999}{1000} \right] = 999$.

The value of $1000 \left[ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{999 \times 1000} \right]$ is

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