If {a_k} = frac{1}{{k(k + 1)}} for k = 1, 2, | vedclass

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(B) We are given ${a_k} = \frac{1}{{k(k + 1)}} = \frac{1}{k} - \frac{1}{{k + 1}}$.Summing from $k = 1$ to $n$:$\sum\limits_{k = 1}^n {{a_k}} = \sum\li

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${\left( {\frac{n}{{n + 1}}} \right)^2}$

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(B) We are given ${a_k} = \frac{1}{{k(k + 1)}} = \frac{1}{k} - \frac{1}{{k + 1}}$.Summing from $k = 1$ to $n$:$\sum\limits_{k = 1}^n {{a_k}} = \sum\limits_{k = 1}^n {\left( {\frac{1}{k} - \frac{1}{{k + 1}}} \right)} $Expanding the sum:$= \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + ... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$This is a telescoping series,so all intermediate terms cancel out:$= 1 - \frac{1}{{n + 1}} = \frac{{n + 1 - 1}}{{n + 1}} = \frac{n}{{n + 1}}$Therefore,${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = {\left( {\frac{n}{{n + 1}}} \right)^2}$.

If ${a_k} = \frac{1}{{k(k + 1)}}$ for $k = 1, 2, 3, 4, ..., n$,then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

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