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Use the formula for the sum of a geometric series to find a power series centered at the origin that converges to the expression. For what values does the series converge? 12/ 3+y The power series 1s Σ Qe. n=0

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Final answer:

To find a power series centered at the origin that converges to the expression 12/(3+y), we use the formula for the sum of a geometric series. The series converges for all values of y between -9 and 9.

Step-by-step explanation:

To find a power series centered at the origin that converges to the expression 12/(3+y), we will use the formula for the sum of a geometric series. The formula for the sum of an infinite geometric series is:

S = a / (1 - r)

Where S represents the sum of the series, a is the first term, and r is the common ratio. In this case, we can rewrite the expression 12/(3+y) as a power of x. Let's assume that x = -y/3. Then, the expression becomes:

12 / (3 + x)

Now, we can see that the first term of our power series is 12 and the common ratio is -x/3. Plugging these values into the formula, we get:

S = 12 / (1 - (-x/3))

Simplifying further, we get:

S = 12 / (1 + x/3)

This is our power series centered at the origin which converges to the expression 12/(3+y). Now, let's determine the values for which the series converges. The series converges when the absolute value of the common ratio, |-x/3| is less than 1. So, |-x/3| < 1. Solving for x, we find that -1 < x/3 < 1. Multiplying by 3, we get -3 < x < 3. Therefore, the series converges when -3 < y/3 < 3. Multiplying by 3 again, we get -9 < y < 9. So, the series converges for all values of y between -9 and 9.

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