If 0 < x < 1, then ${3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....$, is equal to :
Solution
Let $t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty$<br><br>$$ = \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty $$<br><br>$$ = 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + .....\infty } \right)$$<br><br>$= {{2{x^2}} \over {1 - x}} - (\ln (1 - x) - x)$<br><br>$\Rightarrow t = {{2{x^2}} \over {1 - x}} + x - \ln (1 - x)$<br><br>$\Rightarrow t = {{x(1 + x)} \over {1 - x}} - \ln (1 - x)$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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