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If X is Positive, the Sum to Infinity of the Series 1 1 + X − 1 − X ( 1 + X ) 2 + ( 1 − X ) 2 ( 1 + X ) 3 − ( 1 − X ) 3 ( 1 + X ) 4 + . . . . . . I S

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प्रश्न

If x is positive, the sum to infinity of the series \[\frac{1}{1 + x} - \frac{1 - x}{(1 + x )^2} + \frac{(1 - x )^2}{(1 + x )^3} - \frac{(1 - x )^3}{(1 + x )^4} + . . . . . . is\]

विकल्प

  • (a) 1/2

  • (b) 3/4 

  • (c) 1 

  • (d) none of these 

MCQ
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उत्तर

(a) \[\frac{1}{2}\] 

\[\text{ Let } S = \frac{1}{\left( 1 + x \right)} - \frac{\left( 1 - x \right)}{\left( 1 + x \right)^2} + \frac{\left( 1 - x \right)^2}{\left( 1 + x \right)^3} - \frac{\left( 1 - x \right)^3}{\left( 1 + x \right)^4} + . . . \infty \]
\[\text{ It is clear that it is a G . P . with a } = \frac{1}{\left( 1 + x \right)} \text{ and }r = - \frac{\left( 1 - x \right)}{\left( 1 + x \right)} . \]
\[ \therefore S = \frac{a}{\left( 1 - r \right)}\]
\[ \Rightarrow S = \frac{\frac{1}{\left( 1 + x \right)}}{\left[ 1 - \left( - \frac{\left( 1 - x \right)}{\left( 1 + x \right)} \right) \right]}\]
\[ \Rightarrow S = \frac{\frac{1}{\left( 1 + x \right)}}{\left[ 1 + \frac{\left( 1 - x \right)}{\left( 1 + x \right)} \right]}\]
\[ \Rightarrow S = \frac{\frac{1}{\left( 1 + x \right)}}{\left[ \frac{\left( 1 + x \right) + \left( 1 - x \right)}{\left( 1 + x \right)} \right]}\]
\[ \Rightarrow S = \frac{1}{2}\]
\[\] 

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अध्याय 20: Geometric Progression - Exercise 20.8 [पृष्ठ ५८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.8 | Q 17 | पृष्ठ ५८

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