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pashok25 [27]
3 years ago
6

1

Mathematics
1 answer:
Oksanka [162]3 years ago
5 0

Answer:

1/64

Step-by-step explanation:

4^ (-11/3) ÷ 4 ^ (-2/3)

We know a^b ÷a^c = a^(b-c)

4 ^(-11/3 - - 2/3)

4^(-11/3 +2/3)

4^(-9/3)

4^ -3

We know a^-b = 1/a^b

1/4^3

1/64

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Which of the following geometric series converges?
Artist 52 [7]

All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is

S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}

Multiply both sides by <em>r</em> :

rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n

Subtract the latter sum from the first, which eliminates all but the first and last terms:

S_n-rS_n=a-ar^n

Solve for S_n:

(1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}

Then as gets arbitrarily large, the term r^n will converge to 0, leaving us with

S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18

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storchak [24]

Answer: 1/4 fraction .25 decimal

Hope this helps :)

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