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2 years ago

Please help me. these problems

Mathematics

1 answer:

jeyben [28]2 years ago

3 0

Answer:

1st problem:

Converges to 6

2nd problem:

Converges to 504

Step-by-step explanation:

You are comparing to $\sum_{k=1}^{\infty} a_1(r)^{k-1}$

You want the ratio r to be between -1 and 1.

Both of these problem are so that means they both have a sum and the series converges to that sum.

The formula for computing a geometric series in our form is $\frac{a_1}{1-r}$ where $a_1$ is the first term.

The first term of your first series is 3 so your answer will be given by:

$\frac{a_1}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}=6$

The second series has r=1/6 and a_1=420 giving me:

$\frac{420}{1-\frac{1}{6}}=\frac{420}{\frac{5}{6}}=420(\frac{6}{5})=504$ .

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3 years ago

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2 years ago

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7 0

3 years ago

Please help me. these problems​

Please help me. these problems