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Find the sum of the real values of $x$ such that the infinite geometric series $x+\frac{1}{2}x^3+\frac{1}{4}x^5+\frac{1}{8}x^7+\dots$ is equ

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Find the sum of the real values of $x$ such that the infinite geometric series $x+\frac{1}{2}x^3+\frac{1}{4}x^5+\frac{1}{8}x^7+\dots$ is equal to $-12$.

Oct 6, 2017
edited by michaelcai  Oct 6, 2017

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Find the sum of the real values of x such that the infinite geometric series $$x+\frac{1}{2}x^3+\frac{1}{4}x^5+\frac{1}{8}x^7+\dots\text{ is equal to }-12$$

This is the sum of a GP

$$a=x\\ r=0.5x^2\\ S_\infty=\frac{a}{1-r}\;\;where\;\;|r|<1\\ |0.5x^2|<1\\ 0.5x^2<1\\ x^2<2\\ -\sqrt2 \(S_\infty=\frac{x}{1-0.5x^2}\\ -12=\frac{x}{1-0.5x^2}\\ -12(1-0.5x^2)=x \\ -12+6x^2=x \\ 6x^2-x -12=0 \\ 6x^2-9x+8x -12=0 \\ 3x(2x-3)+4(2x-3)=0\\ (3x+4)(2x-3)=0\\ x=-4/3\quad or \quad x=3/2\\ 3/2 >\sqrt2\quad \text{so it is illiminated}\\~\\ \therefore\;\;x=-1\frac{1}{3}$$

check

$$a=-4/3\quad \\ r=0.5*16/9 = 8/9\\ S_\infty=\frac{-4}{3(1-\frac{8}{9})}\\ S_\infty=-12\\$$

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Oct 6, 2017