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The value of the series 4+4x+4x^2+4x^3... is a real number greater than 8. Find all possible values of x, giving your answer in interval notation.

Find $$1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{36} + \dotsb,$$ where we alternately multiply by 1/2 and 1/3 to get the next term.

In a certain infinite geometric series, the first term is 1, and each term is equal to the sum of the two terms after it. (For example, the first term is equal to the sum of the second term and third term). If s is the sum of the series and r is the common ratio, find s - r.

Feb 24, 2019

#1
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Sum of a geometric series   Sn =  a1/(1-r)

in this geometric series   r = x     a1 = 4

if you want total to be greater than 8:

8< 4/(1-x)

8-8x < 4

4<8x

x> 1/2         Interval notation:     (1/2, 1)      we have to limit it to 1    r cannot equal 1 (the equation would have 0 in denominator) and it cannot be GREATER than one....because then the series becomes divergent and has no sum.

Feb 24, 2019
edited by ElectricPavlov  Feb 24, 2019
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Feb 25, 2019
#3
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3.

In a certain infinite geometric series, the first term is 1, and each term is equal to the sum of the two terms after it.

(For example, the first term is equal to the sum of the second term and third term).

If s is the sum of the series and r is the common ratio, find s - r.

$$\text{The infinite geometric series, the first term is 1, is: \large{1+r+r^2+r^3+r^4+\ldots} } \\ \text{The sum of this infinite geometric series is: \large{s=\dfrac{1}{1-r}} } \\ \text{Each term is equal to the sum of the two terms after it:  \large{1=r+r^2} or \mathbf{r^2 = 1-r} }$$

$$\begin{array}{|rcll|} \hline s-r &=& \dfrac{1}{1-r} - r \\\\ s-r &=& \dfrac{1}{1-r} - r\cdot \left(\dfrac{1-r}{1-r}\right) \\\\ s-r &=& \dfrac{1-r(1-r)}{1-r} \\\\ s-r &=& \dfrac{(1-r)+r^2}{1-r} \quad | \quad \mathbf{r^2 = 1-r} \\\\ s-r &=& \dfrac{(1-r)+(1-r)}{1-r} \\\\ s-r &=& \dfrac{2(1-r)}{(1-r)} \\\\ \mathbf{s-r} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}$$

Feb 25, 2019