If a k = k ( k + 1 ) 1 , for k = 1 , 2 , 3 …… . n , then ( ∑ k = 1 n a k ) 2 =

( 1 1 − 2 1 + 2 1 − 3 1 + …… + n 1 − n + n 1 ) 2 ( n + 1 ) 2 n 2

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

MathematicsMediumsingle choice

If $a_{k} = \frac{1}{k ( k + 1 )}$ , for $k = 1, 2, 3 \dots\dots . n$ , then $(\sum_{k = 1}^{n} a_{k})^{2} =$

Answer:

Solution:

$(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots\dots + \frac{1}{n} - \frac{1}{n + n})^{2}$

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

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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