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romanna [79]
3 years ago
12

Find the sum of the geometric sequence. (1 point) 1, 1/2, 1/4, 1/8, 1/16

Mathematics
2 answers:
ASHA 777 [7]3 years ago
8 0
Sn  = a1 * (1 - r^n)
                 ----------
                   1 - r

n = 5   so answer  =

   1 * (1 -  (1/2)^5
        ---------------  =       31/16  
               1 - 1/2
           
wariber [46]3 years ago
8 0
That finite sequence we can just add up, using 16 as a common denominator

1 + 1/2 + 1/4 + 1/8+1/16 = 16/16 + 8/16 + 4/16+2/16+1/16 = (16+8+4+2+1)/16=31/16

They probably want you to use the formula

S_n = \sum_{k=0}^{n-1} r^k = \dfrac{1 - r^n}{1-r}

That's the special case of a geometric series where the first term is one.  If it's not we multiply the answer by the first term.

In our example, r=1/2 and n=5, meaning five terms.

S_5 = \dfrac{ 1- (1/2)^5}{1- (1/2)} = \dfrac{1-1/32}{1/2}=\dfrac{31/32}{1/2}=\dfrac{31}{15}

Same answer, math works!

If the question was asking about the infinite geometric series, that's just a little bigger:

S = \dfrac{1}{1-r} = \dfrac{1}{1 - \frac 1 2} = 2


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\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}

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\frac{-10y}{-10} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \  \frac{10y}{10}

\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \  y

-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{-8-3a}{-10}

\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \  -\frac{-3a-8}{10} \ \textgreater \  y=-\frac{-8-3a}{10}

\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \  6x+10\cdot \frac{8}{10-3a}=20

10\cdot \frac{8}{10-3a} \ \textgreater \  \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{8\cdot \:10}{10-3a}
\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \  \frac{80}{10-3a}

6x+\frac{80}{10-3a}=20 \ \textgreater \  \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}
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20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \  \frac{20}{1}-\frac{80}{-3a+10}

\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10

Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \  \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}
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\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \   \frac{10\left(-a+2\right)}{-3a+10}

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7 0
3 years ago
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