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yarga [219]
3 years ago
5

The sum of 30 times (1/3)^(n-1) from 1 to infinity

Mathematics
1 answer:
sergeinik [125]3 years ago
8 0

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}

and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

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3 0
3 years ago
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Sergeeva-Olga [200]

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3 years ago
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s344n2d4d5 [400]
<h3>Answer: Choice D</h3>

=======================================================

Explanation:

The long way to do this is to multiply all the fractions out by hand, or use a calculator to make shorter work of this.

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The rule is: if there are an even number of negative signs, then the product will be positive. Otherwise, the product is negative.

For choice A, we have 3 negative signs. The result (whatever number it is) is negative. Choice B is a similar story. Choice C is also negative because we have 1 negative sign. Choices A through C have an odd number of negative signs.

Only choice D has an even number of negative signs. The two negatives multiply to cancel to a positive. The negative is like undoing the positive. So two negatives just undo each other. This is why the multiplied version of choice D will be some positive number.

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3 years ago
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Answer:

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