Question 1135462
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*[Tex \LARGE \displaystyle 9 - 18 + 36 - 72 + \ldots = \sum_{n=0}^{\infty}9(-2)^n]
Which is the sum of 9(-2)^n from n = 0 to infinity (aka the terms go on forever)
Therefore the answer is choice D



Note how plugging n = 0 into 9(-2)^n leads to
9(-2)^n = 9(-2)^0
9(-2)^n = 9(1)
9(-2)^n = 9
So the first term to be added is 9


Then trying n = 1 leads to
9(-2)^n = 9(-2)^1
9(-2)^n = 9(-2)
9(-2)^n = -18
Making -18 to be the second term to be added. So far we have the expression *[Tex \LARGE \displaystyle 9 - 18]


Trying n = 2 leads to
9(-2)^n = 9(-2)^2
9(-2)^n = 9(4)
9(-2)^n = 36
We add on 36 as the third term. So far we have the expression *[Tex \LARGE \displaystyle 9 - 18 + 36]


Trying n = 3 leads to
9(-2)^n = 9(-2)^3
9(-2)^n = 9(-8)
9(-2)^n = -72
We add on -72 as the fourth term. So far we have the expression *[Tex \LARGE \displaystyle 9 - 18 + 36 - 72]


As n heads off to infinity, we end up with the expression *[Tex \LARGE \displaystyle 9 - 18 + 36 - 72 + \ldots]


side notes
<ol>
<li> The sequence {9, -18, 36, -72, ...} is geometric.</li>
<li> The starting term is a1 = 9 </li>
<li> The common ratio is r = -2</li>
<li> The common ratio can be found by dividing any term by its previous one (eg: term3/term2 = 36/(-18) = -2)</li>
<li> The infinite series 9-18+36-72+... diverges meaning the sum of the infinitely many terms does not approach a fixed value (ie the sum keeps drastically changing as you add on more terms). The infinite series only converges if -1 < r < 1, but r = -2 is outside this range. </li>
</ol>

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