यदि {a_k} = frac{1}{{k(k + 1)}}, जबकि k = 1,, | vedclass

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Question

$\sum\limits_{k = 1}^n {{a_k}}  = \sum\limits_{k = 1}^n {\frac{1}{{k\,(k + 1)}}} $

=$\left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} -

Vedclass · Accepted Answer

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

Vedclass · Answer

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

यदि ${a_k} = \frac{1}{{k(k + 1)}},$ जबकि $k = 1,\,2,\,3,\,4,.....,\,n$, तब ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

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