The sum of the first 20 terms of the series f | vedclass

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(A) The $n$-th term of the series is given by $T_n = \frac{1}{(n+4)(n+5)}$.We can write this as $T_n = \frac{1}{n+4} - \frac{1}{n+5}$.For the first $2

Vedclass · Accepted Answer

$0.16$

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(A) The $n$-th term of the series is given by $T_n = \frac{1}{(n+4)(n+5)}$.We can write this as $T_n = \frac{1}{n+4} - \frac{1}{n+5}$.For the first $20$ terms,the sum $S_{20} = \sum_{n=1}^{20} \left( \frac{1}{n+4} - \frac{1}{n+5} \right)$.Expanding the sum: $S_{20} = \left( \frac{1}{5} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{8} \right) + \cdots + \left( \frac{1}{24} - \frac{1}{25} \right)$.This is a telescoping series where intermediate terms cancel out.$S_{20} = \frac{1}{5} - \frac{1}{25}$.$S_{20} = \frac{5 - 1}{25} = \frac{4}{25}$.$S_{20} = \frac{16}{100} = 0.16$.

The sum of the first $20$ terms of the series $\frac{1}{5 \times 6} + \frac{1}{6 \times 7} + \frac{1}{7 \times 8} + \cdots$ is

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