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AP Calculus BC Practice Quiz: Representing Functions as Power Series

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Which of the following is a power series representation for the function f(x) = 1 / (1 + 3x)?

All Questions (7)

Which of the following is a power series representation for the function f(x) = 1 / (1 + 3x)?

A) ∑ (from n=0 to ∞) (3x)^n

B) ∑ (from n=0 to ∞) (-3x)^n

C) ∑ (from n=0 to ∞) (-1)^n * 3x^n

D) ∑ (from n=0 to ∞) (3x)^n / n!

Correct Answer: B

The function f(x) = 1 / (1 + 3x) can be written in the form a / (1 - r), which is the sum of a geometric series. Here, a = 1 and r = -3x. The power series is given by the formula ∑ (from n=0 to ∞) ar^n. Substituting the values, we get ∑ (from n=0 to ∞) 1 * (-3x)^n, which is ∑ (from n=0 to ∞) (-3x)^n.

Which of the following is a power series representation for the function f(x) = x^2 / (1 - x^3)?

A) ∑ (from n=0 to ∞) x^(3n)

B) ∑ (from n=0 to ∞) x^(3n+2)

C) ∑ (from n=0 to ∞) x^(5n)

D) ∑ (from n=0 to ∞) x^(n+5)

Correct Answer: B

First, find the power series for 1 / (1 - x^3) using the geometric series formula with r = x^3. This gives ∑ (from n=0 to ∞) (x^3)^n = ∑ (from n=0 to ∞) x^(3n). Then, multiply this series by x^2: f(x) = x^2 * ∑ (from n=0 to ∞) x^(3n) = ∑ (from n=0 to ∞) x^2 * x^(3n) = ∑ (from n=0 to ∞) x^(3n+2).

A power series for the function f(x) = 1 / (1 - x) is given by ∑ (from n=0 to ∞) x^n. Which of the following is a power series for g(x) = 1 / (1 - x)^2?

A) ∑ (from n=1 to ∞) nx^(n-1)

B) ∑ (from n=0 to ∞) x^(n+1) / (n+1)

C) ∑ (from n=0 to ∞) (x^2)^n

D) ∑ (from n=0 to ∞) (-1)^n * nx^(n-1)

Correct Answer: A

The function g(x) = 1 / (1 - x)^2 is the derivative of f(x) = 1 / (1 - x). We can find the power series for g(x) by differentiating the power series for f(x) term-by-term. d/dx [∑ (from n=0 to ∞) x^n] = d/dx [1 + x + x^2 + x^3 + ...] = 0 + 1 + 2x + 3x^2 + ... = ∑ (from n=1 to ∞) nx^(n-1).

Which of the following is a power series representation for the function f(x) = ln(1 + x)?

A) ∑ (from n=0 to ∞) (-1)^n * x^n

B) ∑ (from n=1 to ∞) (-1)^(n-1) * nx^(n-1)

C) ∑ (from n=1 to ∞) (-1)^(n-1) * x^n / n

D) ∑ (from n=0 to ∞) x^n / n!

Correct Answer: C

We know that the derivative of ln(1 + x) is 1 / (1 + x). The power series for 1 / (1 + x) is ∑ (from n=0 to ∞) (-x)^n = ∑ (from n=0 to ∞) (-1)^n * x^n. To find the series for ln(1 + x), we integrate this series term-by-term: ∫ [∑ (from n=0 to ∞) (-1)^n * t^n] dt = C + ∑ (from n=0 to ∞) (-1)^n * x^(n+1) / (n+1). Since ln(1+0) = 0, C=0. The series is ∑ (from n=0 to ∞) (-1)^n * x^(n+1) / (n+1). Re-indexing by letting k = n+1 gives ∑ (from k=1 to ∞) (-1)^(k-1) * x^k / k, which matches option C (with n instead of k).

Which of the following is a power series representation for the function f(x) = 1 / (4 + x^2)?

A) ∑ (from n=0 to ∞) (-1)^n * x^(2n) / 4^n

B) ∑ (from n=0 to ∞) x^(2n) / 4^(n+1)

C) ∑ (from n=0 to ∞) (-1)^n * x^(2n) / 4^(n+1)

D) ∑ (from n=0 to ∞) (-x^2/4)^n

Correct Answer: C

To use the geometric series formula, we must manipulate the function into the form a / (1 - r). f(x) = 1 / (4 + x^2) = 1 / [4 * (1 + x^2/4)] = (1/4) * [1 / (1 - (-x^2/4))]. This is a geometric series with a = 1/4 and r = -x^2/4. The series is (1/4) * ∑ (from n=0 to ∞) (-x^2/4)^n = (1/4) * ∑ (from n=0 to ∞) (-1)^n * x^(2n) / 4^n = ∑ (from n=0 to ∞) (-1)^n * x^(2n) / 4^(n+1).

Which of the following is the Maclaurin series for the function f(x) = arctan(x)?

A) ∑ (from n=0 to ∞) (-1)^n * x^(2n)

B) ∑ (from n=0 to ∞) (-1)^n * x^(2n+1)

C) ∑ (from n=0 to ∞) (-1)^n * x^(2n+1) / (2n+1)

D) ∑ (from n=0 to ∞) x^(2n+1) / (2n+1)

Correct Answer: C

The derivative of arctan(x) is 1 / (1 + x^2). We can find the power series for 1 / (1 + x^2) using the geometric series formula with r = -x^2, which gives ∑ (from n=0 to ∞) (-x^2)^n = ∑ (from n=0 to ∞) (-1)^n * x^(2n). To find the series for arctan(x), we integrate this series term-by-term: ∫ [∑ (from n=0 to ∞) (-1)^n * t^(2n)] dt = C + ∑ (from n=0 to ∞) (-1)^n * x^(2n+1) / (2n+1). Since arctan(0) = 0, the constant of integration C is 0. Thus, the series is ∑ (from n=0 to ∞) (-1)^n * x^(2n+1) / (2n+1).

The function g(x) = 1 / (1 + 4x^2) is represented by the power series ∑ (from n=0 to ∞) (-1)^n * (4x^2)^n. Using this information, which of the following is a power series representation for f(x) = x / (1 + 4x^2)^2?

A) ∑ (from n=1 to ∞) (-1)^(n+1) * n * 4^(n-1) * x^(2n-1)

B) ∑ (from n=1 to ∞) (-1)^n * n * 4^n * x^(2n-1)

C) ∑ (from n=0 to ∞) (-1)^n * 4^n * x^(2n+1)

D) ∑ (from n=0 to ∞) (-1)^n * 4^n * x^(2n+1) / (2n+1)

Correct Answer: A

Notice that f(x) is related to the derivative of g(x). Let's find g'(x). g'(x) = d/dx [ (1 + 4x^2)^(-1) ] = -1 * (1 + 4x^2)^(-2) * (8x) = -8x / (1 + 4x^2)^2. We can see that f(x) = -g'(x) / 8. Now, we differentiate the series for g(x) term-by-term: g'(x) = d/dx [∑ (from n=0 to ∞) (-1)^n * 4^n * x^(2n)] = ∑ (from n=1 to ∞) (-1)^n * 4^n * (2n) * x^(2n-1). Therefore, f(x) = (-1/8) * ∑ (from n=1 to ∞) (-1)^n * 4^n * (2n) * x^(2n-1) = ∑ (from n=1 to ∞) (-1)^(n+1) * (2n/8) * 4^n * x^(2n-1) = ∑ (from n=1 to ∞) (-1)^(n+1) * (n/4) * 4^n * x^(2n-1) = ∑ (from n=1 to ∞) (-1)^(n+1) * n * 4^(n-1) * x^(2n-1).