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lilavasa [31]
3 years ago
11

Find the sum of the following infinite geometric series, if it exists. ½ + -¼ + ⅛ + -1/16 +...

Mathematics
1 answer:
Dominik [7]3 years ago
7 0
Hmm

each term is multiplied by a constant
1/2 times what=-1/4? that is -1/2
-1/4 times what=1/8? that is -1/2

so r=-1/2

does it converge or diverge?
if it converges, it has an infinite sum
if it diverges, it does not

if |r|<1 then it converges
|-1/2|<1?
1/2<1?
true
it converges and has a sum

the sum of an infinite geometric series is
S_n=\frac{a_1}{1-r} where a1 is the first term and r=common ratio

a1=1/2
r=-1/2

S_{\infty}=\frac{\frac{1}{2}}{1-(\frac{-1}{2})}
S_{\infty}=\frac{\frac{1}{2}}{1+\frac{1}{2}}
S_{\infty}=\frac{\frac{1}{2}}{\frac{3}{2}}
S_{\infty}=(\frac{1}{2})(\frac{2}{3})
S_{\infty}=\frac{2}{6}
S_{\infty}=\frac{1}{3}



the sum is \frac{1}{3}
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<img src="https://tex.z-dn.net/?f=x%20%20%5Cdiv%203x%20-%202%20%3D%201%20%5Cdiv%203x%20-%204" id="TexFormula1" title="x \div 3x
Maurinko [17]

Hello from MrBillDoesMath!

Answer:

x = 1/2 (1 +\- i sqrt(23))

Discussion:

x \3x - 2 =   (x/3)*x - 2   =  (x^2)/3 - 2     (*)

1 \3x - 4   =  (1/3)x - 4                               (**)

(*) = (**)   =>

(x^2)/3 -2 = (1/3)x - 4        => multiply both sides by 3

x^2 - 6 = x - 12                 => subtract x from both sides

x^2 -x -6 = -12                   => add 12 to both sides

x^2-x +6  = 0

Using the quadratic formula gives:

x = 1/2 (1 +\- i sqrt(23))

Thank you,

MrB

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