If 0 < x < 1 and $y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....$, then the value of e1 + y at $x = {1 \over 2}$ is :
Solution
$$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$<br><br>$$ = ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$<br><br>$$ = {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ....} \right)$$<br><br>$= {x \over {1 - x}} + \ln (1 - x)$<br><br>$x = {1 \over 2} \Rightarrow y = 1 - \ln 2$<br><br>${e^{1 + y}} = {e^{1 + 1 - \ln 2}}$<br><br>$= {e^{2 - \ln 2}} = {{{e^2}} \over 2}$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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