यदि $x=\frac{2}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{2}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{2}{5}\right)^3+\ldots$ है,तो $x+\frac{1}{x}=$

  • A
    $\frac{1+\sqrt{5}}{4}$
  • B
    $3$
  • C
    $\frac{5 \sqrt{5}+3}{4}$
  • D
    $\frac{5 \sqrt{5}-3}{4}$

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कथन $(A) : 1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{4}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{8}+\ldots \infty = \sqrt[3]{4}$
कारण $(R) : |x| < 1, (1-x)^{-n} = 1+nx+\frac{n(n+1)}{1 \cdot 2} x^2+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^3+\ldots$ सही उत्तर है

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यदि $x = \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \ldots$ है,तो $3x^2 + 6x =$

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