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FrozenT [24]
3 years ago
9

given that the following two are geometric series are convergent: 1+x+x^2+x^3+...and 1-x+x^2-x^3+... determine the value(s) of x

for which the sum of the two series is equal to 8​
Mathematics
1 answer:
DENIUS [597]3 years ago
5 0

Let <em>S</em> and <em>T</em> denote the two finite sums,

<em>S</em> = 1 + <em>x</em> + <em>x</em> ² + <em>x</em> ³ + … + <em>x</em> ᴺ

<em>T</em> = 1 - <em>x</em> + <em>x</em> ² - <em>x</em> ³ + … + (-<em>x</em>) ᴺ

• If both <em>S</em> = 8 and <em>T</em> = 8 as <em>N</em> goes to infinity:

Then

<em>xS</em> = <em>x</em> + <em>x</em> ² + <em>x</em> ³ + <em>x</em> ⁴ + … + <em>x</em> ᴺ⁺¹

-<em>xT</em> = -<em>x</em> + <em>x</em> ² - <em>x</em> ³ + <em>x</em> ⁴ + … + (-<em>x</em>) ᴺ⁺¹

so that

<em>S</em> - <em>xS</em> = 1 - <em>x</em> ᴺ⁺¹   ==>   <em>S</em> = (1 - <em>x</em> ᴺ⁺¹)/(1 - <em>x</em>)

and similarly,

<em>T</em> = (1 - (-<em>x</em>) ᴺ⁺¹)/(1 + <em>x</em>)

For both sums, so long as |<em>x</em>| < 1, we have

lim [<em>N</em> → ∞] <em>S</em> = 1/(1 - <em>x</em>)

lim [<em>N</em> → ∞] <em>T</em> = 1/(1 + <em>x</em>)

Then if both sums converge to 8, this happens for

<em>S </em>: 1/(1 - <em>x</em>) = 8   ==>   <em>x</em> = 7/8

<em>T</em> : 1/(1 + <em>x</em>) = 8   ==>   <em>x</em> = -7/8

• If the sum <em>S</em> + <em>T</em> = 8 as <em>N</em> goes to infinity:

From the previous results, we have

1/(1 - <em>x</em>) + 1/(1 + <em>x</em>) = 8   ==>   <em>x</em> = ±√3/2

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