AP Calculus BC Quiz: Taylor And Maclaurin Series

What this quiz covers

This quiz focuses on Taylor And Maclaurin Series, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{(x-1)^n}{n}$ (correct answer)
2. $\sum_{n=0}^{\infty}(-1)^{n}\dfrac{(x-1)^n}{n}$
3. $\sum_{n=1}^{\infty}(-1)^{n}\dfrac{(x-1)^n}{n}$
4. $\sum_{n=1}^{\infty}\dfrac{(x-1)^n}{n!}$
5. $\sum_{n=0}^{\infty}(-1)^{n-1}\dfrac{(x-1)^{n+1}}{(n+1)!}$

Explanation: This problem asks for the Taylor series of $F(x) = \ln x$ centered at $x = 1$, a classic Taylor series application. We compute derivatives: $F'(x) = \frac{1}{x}$, $F''(x) = -\frac{1}{x^2}$, $F'''(x) = \frac{2}{x^3}$, and evaluate at $x = 1$ to get $F^{(n)}(1) = (-1)^{n-1}(n-1)!$ for $n \geq 1$. The Taylor series formula gives $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(x-1)^n}{n}$, noting that $F(1) = \ln(1) = 0$ so the series starts at $n = 1$. Option C has the wrong sign pattern with $(-1)^n$ instead of $(-1)^{n-1}$, which would make the first term negative. For Taylor series centered at $a$, always use powers of $(x-a)$ and evaluate derivatives at $x = a$.

Question 2

1. $\sum_{n=0}^{\infty}\dfrac{(2x)^n}{n!}$ (correct answer)
2. $\sum_{n=0}^{\infty}\dfrac{2^n x^n}{(n+1)!}$
3. $\sum_{n=0}^{\infty}\dfrac{(2x)^{n+1}}{n!}$
4. $\sum_{n=0}^{\infty}(-1)^n\dfrac{(2x)^n}{n!}$
5. $\sum_{n=0}^{\infty}\dfrac{(2x)^n}{(2n)!}$

Explanation: This problem asks for the Maclaurin series of $T(x) = e^{2x}$, which requires applying the Taylor/Maclaurin series expansion technique. The Maclaurin series for $e^u$ is $\sum_{n=0}^{\infty}\frac{u^n}{n!}$, so substituting $u = 2x$ gives us $\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}$. Each term expands as $\frac{(2x)^n}{n!} = \frac{2^n x^n}{n!}$, confirming that we need $(2x)^n$ in the numerator. Option B incorrectly uses $(n+1)!$ in the denominator instead of $n!$, which would shift all the factorial terms. The key strategy is to substitute the composite argument directly into the known series formula for the base function.

Question 3

1. $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{2n+1}$ (correct answer)
2. $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{2n+1}$
3. $\sum_{n=0}^{\infty}(-1)^{n+1}\dfrac{x^{2n+1}}{2n+1}$
4. $\sum_{n=0}^{\infty}\dfrac{x^{2n+1}}{(2n+1)!}$
5. $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$

Explanation: Finding the Maclaurin series for $q(x) = \arctan(x)$ requires the Taylor/Maclaurin series approach through integration. Since $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$, integrating term by term gives $\arctan(x) = \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}$. The series has odd powers only, with alternating signs and denominators that are odd integers. Option E incorrectly uses factorials $(2n+1)!$ in the denominator instead of just $(2n+1)$, which would make the series converge much faster than the actual arctangent series. Integration of a power series increases the exponent by 1 and divides by the new exponent, not by a factorial.

Question 4

1. $1+x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+\dfrac{x^5}{5!}-\dfrac{x^6}{6!}$
2. $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}$ (correct answer)
3. $1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}$
4. $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}$
5. $1-\dfrac{x^2}{2}+\dfrac{x^4}{6}-\dfrac{x^6}{24}$

Explanation: This problem involves finding the Maclaurin series expansion for the function q(x) = cos x. To construct the series, compute derivatives at 0: cos(0)=1, -sin(0)=0, -cos(0)=-1, sin(0)=0, cos(0)=1, etc., showing even powers with alternating signs. Dividing by n! yields 1 - x^2/2! + x^4/4! - x^6/6!. Odd derivatives at 0 are zero, so only even terms remain. A tempting distractor like choice C fails because it uses all positive signs, neglecting the alternating pattern from cosine's higher derivatives. A transferable series-construction strategy is to remember cosine's series mirrors sine's but shifted to even powers starting with 1.

Question 5

1. $1+x+x^2+x^3+x^4+x^5$
2. $1-x+x^2-x^3+x^4-x^5$ (correct answer)
3. $1-x-x^2-x^3-x^4-x^5$
4. $1-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\dfrac{x^4}{5}-\dfrac{x^5}{6}$
5. $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$

Explanation: This problem involves finding the Maclaurin series expansion for the function m(x) = 1/(1 + x). To construct the series, recognize it as the geometric series sum_{n=0}^∞ (-1)^n x^n, converging for |x| < 1. Through x^5, this is 1 - x + x^2 - x^3 + x^4 - x^5. This is derived by factoring out the negative sign compared to 1/(1 - (-x)). A tempting distractor like choice A fails because it uses all positive signs, which applies to 1/(1 - x) instead. A transferable series-construction strategy is to adjust the geometric series for the sign in the denominator to incorporate alternating terms.

Question 6

1. $(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$ (correct answer)
2. $1+(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$
3. $(x-1)+\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}$
4. $(x-1)-\dfrac{(x-1)^2}{3}+\dfrac{(x-1)^3}{4}$
5. $(x-1)-\dfrac{(x-1)^2}{2!}+\dfrac{(x-1)^3}{3!}$

Explanation: This problem involves finding the Taylor series expansion for $f(x) = \ln x$ centered at $x=1$. To construct the series, compute derivatives at 1: $f(1)=0$, $f'(1)=1$, $f''(1)=-1$, $f'''(1)=2$, and divide by n!: $(x-1) - \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3}$ through the cubic term. This follows the pattern for $\ln(1 + u)$ with $u = x-1$. The constant term is zero since $\ln(1)=0$. A tempting distractor like choice B fails because it includes an unnecessary +1 constant term, which would be incorrect as the series starts from the linear term. A transferable series-construction strategy is to substitute a shift like $u = x - a$ into a known Maclaurin series to center it at a point $a$.

Question 7

1. $\displaystyle \sum_{n=0}^{\infty} \frac{(-2x)^n}{n}$
2. $\displaystyle \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!}$ (correct answer)
3. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{2x^n}{n!}$
4. $\displaystyle \sum_{n=1}^{\infty} \frac{(-2x)^n}{n!}$
5. $\displaystyle \sum_{n=0}^{\infty} \frac{(2x)^n}{n!}$

Explanation: This question examines Maclaurin series, which are Taylor expansions at x=0. The exponential function e^u has the series ∑ u^n / n! from n=0 to ∞. Substitute u = -2x to get e^{-2x} = ∑ (-2x)^n / n! from n=0 to ∞. This series converges for all x. A tempting distractor is choice E, which uses (2x)^n instead of (-2x)^n, but this represents e^{2x}, failing due to the positive exponent. A useful strategy for series construction is to substitute linear terms into well-known exponential or trigonometric series.

Question 8

1. $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n}$
2. $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$ (correct answer)
3. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{n}}{n+1}$
4. $\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{x^n}{n!}$
5. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$

Explanation: This question tests your ability to find the Maclaurin series for ln(1+x). To construct this series, we start by recognizing that the derivative of ln(1+x) is 1/(1+x), which has the geometric series representation 1-x+x²-x³+... = Σ(-1)ⁿxⁿ for |x|<1. Integrating term by term gives us ∫(1-x+x²-x³+...)dx = x-x²/2+x³/3-x⁴/4+... = Σ(-1)ⁿ⁺¹xⁿ/n, starting from n=1. Choice A incorrectly omits the alternating signs that come from the geometric series of 1/(1+x). The key strategy is to recognize when you can use the derivative or integral of a known series to find a new series.

Question 9

1. $\sum_{n=0}^{\infty}(-1)^n\dfrac{(3x)^{2n+1}}{(2n+1)!}$ (correct answer)
2. $\sum_{n=0}^{\infty}(-1)^n\dfrac{(3x)^{2n}}{(2n)!}$
3. $\sum_{n=0}^{\infty}(-1)^n\dfrac{(3x)^{2n+1}}{(2n)!}$
4. $\sum_{n=0}^{\infty}(-1)^{n+1}\dfrac{(3x)^{2n+1}}{(2n+1)!}$
5. $\sum_{n=0}^{\infty}\dfrac{(3x)^{2n+1}}{(2n+1)!}$

Explanation: To find the Maclaurin series for $s(x) = \sin(3x)$, we apply the Taylor/Maclaurin series expansion method. The Maclaurin series for $\sin(u)$ is $\sum_{n=0}^{\infty}(-1)^n\frac{u^{2n+1}}{(2n+1)!}$, so substituting $u = 3x$ yields $\sum_{n=0}^{\infty}(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}$. This expansion captures the alternating signs and odd powers characteristic of the sine function. Option E temptingly omits the $(-1)^n$ factor, which would incorrectly make all terms positive and fail to represent the oscillating nature of sine. Remember that sine series always have odd powers with alternating signs, while cosine series have even powers with alternating signs.

Question 10

1. $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}$ (correct answer)
2. $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$
3. $\sum_{n=0}^{\infty}(-1)^{n+1}\dfrac{x^{2n}}{(2n)!}$
4. $\sum_{n=0}^{\infty}\dfrac{x^{2n}}{(2n+1)!}$
5. $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{n!}$

Explanation: Finding the Maclaurin series for $h(x) = \cos(x)$ is a fundamental Taylor/Maclaurin series problem. The cosine function has derivatives that cycle through $\cos(x)$, $-\sin(x)$, $-\cos(x)$, $\sin(x)$, and evaluating at $x=0$ gives the pattern $1, 0, -1, 0, 1, ...$. This produces the series $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$, containing only even powers of $x$ with alternating signs. Option B incorrectly includes odd powers, which would appear in the sine series but not cosine. Remember that cosine is an even function, so its Maclaurin series contains only even powers of $x$.

Question 11

1. $\sum_{n=0}^{\infty}(2x)^n$ (correct answer)
2. $\sum_{n=1}^{\infty}(2x)^n$
3. $\sum_{n=0}^{\infty}(-2x)^n$
4. $\sum_{n=0}^{\infty}\dfrac{(2x)^n}{n!}$
5. $\sum_{n=0}^{\infty}\dfrac{(2x)^{n+1}}{n+1}$

Explanation: This problem asks for the Maclaurin series of $g(x) = \frac{1}{1-2x}$, which is a classic application of Taylor/Maclaurin series. The geometric series formula states that $\frac{1}{1-u} = \sum_{n=0}^{\infty}u^n$ for $|u| < 1$, so substituting $u = 2x$ gives $\sum_{n=0}^{\infty}(2x)^n$. This series converges when $|2x| < 1$, or $|x| < \frac{1}{2}$. Option C incorrectly uses $(-2x)^n$, which would represent $\frac{1}{1+2x}$ instead of $\frac{1}{1-2x}$. The key insight is recognizing when a function matches the geometric series pattern $\frac{1}{1-u}$ and making the appropriate substitution.

Question 12

1. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{4n}}{(2n)!}$ (correct answer)
2. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!}$
3. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{4n}}{(4n)!}$
4. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{4n+1}}{(2n)!}$
5. $\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^{4n}}{(2n)!}$

Explanation: This question assesses Maclaurin series, centered at zero. The cosine function is cos(u) = ∑ (-1)^n u^{2n} / (2n)! from n=0 to ∞. Substitute u = x^2 to get cos(x^2) = ∑ (-1)^n (x^2)^{2n} / (2n)! = ∑ (-1)^n x^{4n} / (2n)! from n=0 to ∞. This holds for all x. A tempting distractor is choice B, with x^{2n} instead of x^{4n}, but this would be cos(x) not cos(x^2), failing the substitution of the argument. An effective strategy is to compose known series by replacing the variable with a polynomial expression like x^2.

Question 13

1. $1+2x+\dfrac{(2x)^2}{2!}+\dfrac{(2x)^3}{3!}+\dfrac{(2x)^4}{4!}$ (correct answer)
2. $1+2x+\dfrac{2x^2}{2!}+\dfrac{2x^3}{3!}+\dfrac{2x^4}{4!}$
3. $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}$
4. $1+2x+\dfrac{(2x)^2}{3!}+\dfrac{(2x)^3}{4!}+\dfrac{(2x)^4}{5!}$
5. $1-2x+\dfrac{(2x)^2}{2!}-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^4}{4!}$

Explanation: This problem involves finding the Maclaurin series expansion for the function g(x) = e^{2x}. To construct the series, start with the known Maclaurin series for e^u = sum u^n / n! and substitute u = 2x. This gives sum (2x)^n / n! = 1 + 2x + (4x^2)/2! + (8x^3)/3! + (16x^4)/4! through the x^4 term. Simplifying, it becomes 1 + 2x + (2^2 x^2)/2 + (2^3 x^3)/6 + (2^4 x^4)/24, which matches the expanded form. A tempting distractor like choice B fails because it incorrectly places the 2 multiplier only outside the powers, resulting in terms like 2x^2 / 2! instead of (2x)^2 / 2!. A transferable series-construction strategy is to use substitution into known series like e^x and carefully apply the chain rule for composite functions.

Question 14

1. $x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}$
2. $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}$ (correct answer)
3. $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}$
4. $x-\dfrac{x^3}{2!}+\dfrac{x^5}{4!}-\dfrac{x^7}{6!}$
5. $x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+\dfrac{x^5}{5!}$

Explanation: This problem involves finding the Maclaurin series expansion for the function h(x) = sin x. To construct the series, use the Taylor series formula centered at 0, computing derivatives: sin(0)=0, cos(0)=1, -sin(0)=0, -cos(0)=-1, sin(0)=0, cos(0)=1, and so on, showing odd powers with alternating signs. Dividing by n! gives terms x/1! - x^3/3! + x^5/5! - x^7/7!, but since 1!=1, it's x - x^3/6 + x^5/120 - x^7/5040. The even derivatives at 0 are zero, so only odd terms appear. A tempting distractor like choice A fails because it uses positive signs for all terms, ignoring the alternating pattern from the sine function's derivatives. A transferable series-construction strategy is to recall the patterns for trigonometric functions, where sine has odd powers and alternating signs starting positive.

Question 15

1. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}$ (correct answer)
2. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{(3x)^{2n}}{(2n)!}$
3. $\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(3x)^{2n+1}}{(2n+1)!}$
4. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{3x^{2n+1}}{(2n+1)!}$
5. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{(3x)^{2n+1}}{(2n)!}$

Explanation: This question tests proficiency in Maclaurin series, expansions centered at zero. The sine function has a known Maclaurin series: $\sin(u) = \sum_{n=0}^{\infty} (-1)^n \frac{u^{2n+1}}{(2n+1)!}$. Substitute u = 3x to obtain $\sin(3x) = \sum_{n=0}^{\infty} (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!}$. This expansion is valid for all x since the sine series converges everywhere. A tempting distractor is choice C, which includes an extra (-1) factor as (-1)^{n+1}, but this would negate the series to $-\sin(3x)$, failing to match the positive leading term. A transferable strategy is to use composition by substituting into known series for elementary functions like sine or cosine.

Question 16

1. $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n}$ (correct answer)
2. $\displaystyle \sum_{n=1}^{\infty} (-1)^{n}\frac{(x-1)^n}{n}$
3. $\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n+1}$
4. $\displaystyle \sum_{n=1}^{\infty} \frac{(x-1)^n}{n}$
5. $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n!}$

Explanation: To find the Taylor series for ln(x) centered at x=1, we use the fact that d/dx[ln(x)] = 1/x. From the previous problem, we know 1/x = Σ(-1)ⁿ(x-1)ⁿ for |x-1|<1. Integrating term by term: ∫[1 - (x-1) + (x-1)² - ...]dx = (x-1) - (x-1)²/2 + (x-1)³/3 - ... = Σ(-1)ⁿ⁺¹(x-1)ⁿ/n, starting from n=1. Choice B has incorrect signs (-1)ⁿ instead of (-1)ⁿ⁺¹, which would give -ln(x) instead of ln(x). The strategy is to differentiate or integrate known series to find related series.

Question 17

1. $1-x+x^2-x^3+x^4-x^5$
2. $1+x+x^2+x^3+x^4+x^5$ (correct answer)
3. $1+x-x^2+x^3-x^4+x^5$
4. $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}$
5. $1+x^2+x^4$

Explanation: This problem involves finding the Maclaurin series expansion for the function p(x) = 1/(1 - x). To construct the series, recognize it as the geometric series sum_{n=0}^∞ x^n, which converges for |x| < 1. Through x^5, this is explicitly 1 + x + x^2 + x^3 + x^4 + x^5. This can be derived by long division or noting the infinite sum formula. A tempting distractor like choice A fails because it alternates signs, which would be correct for 1/(1 + x) instead of 1/(1 - x). A transferable series-construction strategy is to identify geometric forms and adjust the sign in the denominator to determine if signs alternate.

Question 18

1. $x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}$
2. $x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}$ (correct answer)
3. $x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}-\dfrac{x^6}{6}+\dfrac{x^7}{7}$
4. $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}$
5. $1-\dfrac{x^2}{3}+\dfrac{x^4}{5}-\dfrac{x^6}{7}$

Explanation: This problem involves finding the Maclaurin series expansion for the function r(x) = arctan x. To construct the series, integrate the geometric series for 1/(1 + x^2) = sum (-1)^n x^{2n}, yielding arctan x = sum (-1)^n x^{2n+1} / (2n+1). Through x^7, this is x - x^3/3 + x^5/5 - x^7/7. Derivatives confirm: r'(x)=1/(1+x^2), and higher terms follow the pattern. A tempting distractor like choice D fails because it uses factorial denominators, confusing it with trigonometric series like sin x. A transferable series-construction strategy is to derive inverse trig series by integrating known geometric or binomial expansions.

Question 19

1. $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}$
2. $x+x^2+\dfrac{x^3}{2!}+\dfrac{x^4}{3!}+\dfrac{x^5}{4!}$ (correct answer)
3. $x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}$
4. $x+x^2+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}$
5. $x-x^2+\dfrac{x^3}{2!}-\dfrac{x^4}{3!}+\dfrac{x^5}{4!}$

Explanation: This problem involves finding the Maclaurin series expansion for the function t(x) = x e^x. To construct the series, multiply the series for e^x = sum x^n / n! by x, giving sum x^{n+1} / n! from n=0. This shifts to x + x^2 / 1! + x^3 / 2! + x^4 / 3! + x^5 / 4!. Equivalently, it's sum_{k=1} x^k / (k-1)!, with coefficients 1, 1, 1/2, 1/6, 1/24. A tempting distractor like choice A fails because it includes a constant term of 1, which is absent since t(0)=0. A transferable series-construction strategy is to manipulate known series through multiplication, differentiation, or integration to build compound functions.

Question 20

1. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$ (correct answer)
2. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{2n+1}$
3. $\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$
4. $\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$
5. $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$

Explanation: This question focuses on Maclaurin series, power series centered at zero. The arctangent function arctan(x) can be derived by integrating the geometric series for 1/(1+t^2). Specifically, 1/(1+x^2) = ∑ (-1)^n x^{2n} from n=0 to ∞ for |x|<1, and integrating gives arctan(x) = ∑ (-1)^n x^{2n+1} / (2n+1) from n=0 to ∞. The constant is zero since arctan(0)=0. A tempting distractor is choice D, which has (-1)^{n+1} instead, but this negates the series to -arctan(x), failing for positive x values. A broad strategy is to derive series for integrals of functions by term-by-term integration of known expansions.