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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
If $0\,<\,x\,<\,1$, then $\frac{3}{2} x^{2}+\frac{5}{3} x^{3}+\frac{7}{4} x^{4}+\ldots . .$, is equal to :
  • ✓
    $\mathrm{x}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)-\log _{\mathrm{e}}(1-\mathrm{x})$
  • B
    $\mathrm{x}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)+\log _{\mathrm{e}}(1-\mathrm{x})$
  • C
    $\frac{1-x}{1+x}+\log _{e}(1-x)$
  • D
    $\frac{1+x}{1-x}+\log _{e}(1-x)$
✓

Answer

Correct option: A.
$\mathrm{x}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)-\log _{\mathrm{e}}(1-\mathrm{x})$
a
Let $t=\frac{3}{2} x^{2}+\frac{5}{3} x^{3}+\frac{7}{4} x^{4}+\ldots . \infty$

$=\left(2-\frac{1}{2}\right) x^{2}+\left(2-\frac{1}{3}\right) x^{3}+\left(2-\frac{1}{4}\right) x^{4}+\ldots \infty$

$=2\left(\mathrm{x}^{2}+\mathrm{x}^{3}+\mathrm{x}^{4}+\ldots \infty\right)-\left(\frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{x}^{4}}{4}+\ldots \infty\right)$

$=\frac{2 x^{2}}{1-x}-(\ell n(1-x)-x)$

$\Rightarrow t=\frac{2 x^{2}}{1-x}+x-\ell n(1-x)$

$\Rightarrow t=\frac{x(1+x)}{1-x}-\ell n(1-x)$

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