If U_{n} = frac{1}{n} - frac{1}{n+1},then the | vedclass

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(B) Given $U_{n} = \frac{1}{n} - \frac{1}{n+1}$.This is a telescoping series.The sum $S_{5} = U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ can be written as

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$\frac{5}{6}$

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(B) Given $U_{n} = \frac{1}{n} - \frac{1}{n+1}$.This is a telescoping series.The sum $S_{5} = U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ can be written as:$S_{5} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{6})$Observing the terms,the intermediate fractions cancel out:$S_{5} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6}$$S_{5} = 1 - \frac{1}{6}$$S_{5} = \frac{6-1}{6} = \frac{5}{6}$

If $U_{n} = \frac{1}{n} - \frac{1}{n+1}$,then the value of $U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ is

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