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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2016 · Mathematics · Sequences and Series

(x+(1)/(x))^ 2 +(x^ 2 + 1 x^ 2 )^ 2 +(x^ 3 + 1 x^ 3 )^ 2 . upto n...

(x+1x)2+(x2+1x2)2+(x3+1x3)2….\left(x+\frac{1}{x}\right)^{2}+\left(x^{2}+\frac{1}{x^{2}}\right)^{2}+\left(x^{3}+\frac{1}{x^{3}}\right)^{2} \quad \ldots . upto nn terms is

Options

  1. Ax2n−1x2−1×x2n+2+1x2n+2n\frac{x^{2 n}-1}{x^{2}-1} \times \frac{x^{2 n+2}+1}{x^{2 n}}+2 n
  2. Bx2n+1x2+1×x2n+2−1x2n−2n\frac{x^{2 n}+1}{x^{2}+1} \times \frac{x^{2 n+2}-1}{x^{2 n}}-2 n
  3. Cx2n−1x2−1×x2n−1x2n−2n\frac{x^{2 n}-1}{x^{2}-1} \times \frac{x^{2 n}-1}{x^{2 n}}-2 n
  4. DNone of these

Correct answer

A. x2n−1x2−1×x2n+2+1x2n+2n\frac{x^{2 n}-1}{x^{2}-1} \times \frac{x^{2 n+2}+1}{x^{2 n}}+2 n

Step-by-step solution

The series is (x2+x4+x6+…)+(1x2+1x4+1x6+…)+(2+2+…..)=x2(x2n−1)x2−1+1x2(1−1x2n)1−1x2+2n=x2(x2n−1)x2−1+x2n−1(x2−1)x2n+2n=x2n−1x2−1×x2n+2+1x2n+2n\begin{array}{l}\left(x^{2}+x^{4}+x^{6}+\ldots\right)+\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}+\frac{1}{x^{6}}+\ldots\right)\\ +(2+2+\ldots . .) \\ =\frac{x^{2}\left(x^{2 n}-1\right)}{x^{2}-1}+\frac{\frac{1}{x^{2}}\left(1-\frac{1}{x^{2 n}}\right)}{1-\frac{1}{x^{2}}}+2 n \\ =\frac{x^{2}\left(x^{2 n}-1\right)}{x^{2}-1}+\frac{x^{2 n}-1}{\left(x^{2}-1\right) 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 \end{array}
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