( x + x 1 ) 2 + ( x 2 + x 2 1 ) 2 + ( x 3 + x 3 1 ) 2 … . upto n terms is

The series is ( x 2 + x 4 + x 6 + … ) + ( x 2 1 + x 4 1 + x 6 1 + … ) + ( 2 + 2 + … .. ) = x 2 − 1 x 2 ( x 2 n − 1 ) + 1 − x 2 1 x 2 1 ( 1 − x 2 n 1 ) + 2 n = x 2 − 1 x 2 ( x 2 n − 1 ) + ( x 2 − 1 ) x 2 n x 2 n − 1 + 2 n = x 2 − 1 x 2 n − 1 × x 2 n x 2 n + 2 + 1 + 2 n

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesHard2 minPYQ_2016

MathematicsHardsingle choice

$(x + \frac{1}{x})^{2} + (x^{2} + \frac{1}{x ^{2}})^{2} +$ $(x^{3} + \frac{1}{x ^{3}})^{2} \dots .$ upto $n$ terms is

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Answer:

Solution:

The series is $(x^{2} + x^{4} + x^{6} + \dots) + (\frac{1}{x ^{2}} + \frac{1}{x ^{4}} + \frac{1}{x ^{6}} + \dots) + (2 + 2 + \dots ..) = \frac{x ^{2} ( x ^{2 n} - 1 )}{x ^{2} - 1} + \frac{\frac{1}{x ^{2}} ( 1 - \frac{1}{x ^{2 n}} )}{1 - \frac{1}{x ^{2}}} + 2 n = \frac{x ^{2} ( x ^{2 n} - 1 )}{x ^{2} - 1} + \frac{x ^{2 n} - 1}{( x ^{2} - 1 ) x ^{2 n}} + 2 n = \frac{x ^{2 n} - 1}{x ^{2} - 1} \times \frac{x ^{2 n + 2} + 1}{x ^{2 n}} + 2 n$

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

⏱ 2mℹ️ Source: PYQ_2016

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