left[frac{1}{1 times 2}+frac{1}{2 times 3}+fr | vedclass

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Question

(B) The general term of the series is given by $\frac{1}{n(n+1)}$.Using partial fractions,we can write $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}

Vedclass · Accepted Answer

$\frac{99}{100}$

Vedclass · Answer

(B) The general term of the series is given by $\frac{1}{n(n+1)}$.Using partial fractions,we can write $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.Applying this to each term:$\frac{1}{1 	imes 2} = \frac{1}{1} - \frac{1}{2}$$\frac{1}{2 	imes 3} = \frac{1}{2} - \frac{1}{3}$$\frac{1}{3 	imes 4} = \frac{1}{3} - \frac{1}{4}$...$\frac{1}{99 	imes 100} = \frac{1}{99} - \frac{1}{100}$Summing these terms:$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \cdots + (\frac{1}{99} - \frac{1}{100})$All intermediate terms cancel out,leaving only the first and the last term:$= 1 - \frac{1}{100} = \frac{100-1}{100} = \frac{99}{100}$.

$\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{99 \times 100}\right]$ is equal to

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