Exact value: [tex]c = \frac{-11+\sqrt{165}}{22}[/tex]
Approximate value: c = 0.08387
Round the approximate value however you need to
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Work Shown:
Let x = 1+c
[tex]\displaystyle S = \sum_{n=2}^{\infty}(1+c)^{-n}\\\\\\\displaystyle S = \sum_{n=2}^{\infty}x^{-n}\\\\\\\displaystyle S = \sum_{n=2}^{\infty}\frac{1}{x^n}\\\\\\\displaystyle S = \frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}\ldots\\\\\\[/tex]
We have an infinite geometric series here. The first term is a = 1/(x^2). The common ratio is r = 1/x.
Each new term is found by multiplying the previous term by 1/x.
Assuming -1 < r < 1 is true, the infinite geometric sum is
[tex]S = \frac{a}{1-r}\\\\\\S = \frac{1/x^2}{1-1/x}\\\\\\S = \frac{1/x^2}{x/x-1/x}\\\\\\S = \frac{1/x^2}{(x-1)/x}\\\\\\S = \frac{1}{x^2}\div\frac{x-1}{x}\\\\\\S = \frac{1}{x^2}*\frac{x}{x-1}\\\\\\S = \frac{1}{x^2-x}\\\\\\[/tex]
Plug in S = 11 and solve for x
[tex]S = \frac{1}{x^2-x}\\\\11 = \frac{1}{x^2-x}\\\\11(x^2-x) = 1\\\\11x^2-11x = 1\\\\11x^2-11x-1 = 0\\\\[/tex]
Use the quadratic formula to find the two solutions
[tex]x = \frac{11+\sqrt{165}}{22} \approx 1.08387\\\\x = \frac{11-\sqrt{165}}{22} \approx -0.08387\\\\[/tex]
Using these x values, we find that the corresponding r values are
r = 1/x = 1/(1.08387) = 0.92262
r = 1/x = 1/(-0.08387) = -11.92321
The first r value makes -1 < r < 1 true, but the second r value does not. So we will be ignoring the solution x = -0.08387
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Using the solution that corresponds to x = 1.08387, we find the value of c is
[tex]x = c+1\\\\c = x-1\\\\c = \frac{11+\sqrt{165}}{22}-1\\\\c = \frac{11+\sqrt{165}}{22}-\frac{22}{22}\\\\c = \frac{11+\sqrt{165}-22}{22}\\\\c = \frac{-11+\sqrt{165}}{22}\\\\c \approx 0.08387\\\\[/tex]
