AP Calculus BC Quiz: Representing Series As Power Series

What this quiz covers

This quiz focuses on Representing Series As Power 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. $\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{n}$
2. $\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{n+1}$ (correct answer)
3. $\displaystyle \sum_{n=1}^{\infty}(-1)^n x^{n}$
4. $\displaystyle \sum_{n=0}^{\infty}(-1)^{n-1} x^{n+1}$
5. $\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{2n+1}$

Explanation: Representing functions as power series is a key skill in AP Calculus BC, essential for financial models involving rational functions. For q(x) = x/(1+x), express it as x * 1/(1 - (-x)). Apply the geometric series sum_{n=0}^∞ (-x)^n = 1/(1 - (-x)), then multiply by x to get sum_{n=0}^∞ (-1)^n x^{n+1}. This aligns with choice B. Choice A is tempting but represents 1/(1+x) without the x multiplier, omitting the initial power shift. A transferable strategy is to factor out terms like polynomials and combine with geometric series, reindexing as needed for the correct powers.

Question 2

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, allowing us to express functions like rational ones as infinite sums for analysis. To derive the series for f(x) = 1/(1-3x), start with the geometric series sum_{n=0}^∞ r^n = 1/(1-r) for |r|<1. Replace r with 3x, yielding sum_{n=0}^∞ (3x)^n = sum_{n=0}^∞ 3^n x^n = 1/(1-3x) for |3x|<1 or |x|<1/3. This matches choice D directly. A tempting distractor like choice C fails because it uses (-3)^n x^n, which would represent 1/(1+3x) instead of 1/(1-3x), altering the sign and thus the function. Remember, a transferable strategy is to manipulate known series like the geometric one by substitution and algebraic adjustments to match the target function.

Question 3

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

Explanation: This question tests the skill of representing functions as power series, using the cosine expansion. The Maclaurin series for cos(y) is $\sum_{n=0}^{\infty} (-1)^n \frac{y^{2n}}{(2n)!}$ for all real y. Replacing y with 2x gives $\sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n} x^{2n}}{(2n)!}$, but choice A keeps it as $(2x)^{2n}$. This is equivalent and converges everywhere. A tempting distractor is choice B, which has odd powers and represents sin(2x) instead. A transferable strategy for representing functions as power series is to substitute into standard Taylor series for transcendental functions like cosine.

Question 4

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

Explanation: This problem asks for the power series of r(x) = 2/(1-4x). We factor out the constant: r(x) = 2·1/(1-4x), where the geometric series gives 1/(1-4x) = Σ(n=0 to ∞) (4x)^n = Σ(n=0 to ∞) 4^n x^n. Therefore, r(x) = 2·Σ(n=0 to ∞) 4^n x^n = Σ(n=0 to ∞) 2·4^n x^n. Choice A would have 2·4^(n+1) = 2·4·4^n = 8·4^n as coefficients, which is four times too large. When constants appear in the numerator, factor them out and multiply the entire series by that constant.

Question 5

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, enabling approximation and integration of functions like logarithms. The series for ln(1+x) derives from integrating the geometric series for 1/(1+t) = sum_{n=0}^∞ (-1)^n t^n for |t|<1. Integrating from 0 to x gives ln(1+x) = integral sum_{n=0}^∞ (-1)^n t^n dt = sum_{n=0}^∞ (-1)^n x^{n+1}/(n+1), then reindexing to n=1 yields sum_{n=1}^∞ (-1)^{n-1} x^n / n. This corresponds to choice A. Choice C is a common distractor but fails as it has (-1)^n instead of (-1)^{n-1}, flipping the signs and representing -ln(1+x). A transferable strategy is to derive new series by integrating or differentiating known ones, adjusting indices accordingly for the desired function.

Question 6

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

Explanation: This problem asks for the power series of $g(x) = \frac{5}{1+2x}$. We can rewrite this as $5 \cdot \frac{1}{1 - (-2x)}$, matching the geometric series form $\frac{1}{1-r} = \sum r^n$. With $r = -2x$, we get $g(x) = 5 \cdot \sum_{n=0}^{\infty} (-2x)^n = 5 \cdot \sum_{n=0}^{\infty} (-2)^n x^n = \sum_{n=0}^{\infty} 5(-2)^n x^n$. Choice A incorrectly writes $(2x)^n$ instead of $(-2x)^n$, missing the alternating signs from the $+2x$ in the denominator. When the denominator has $1 + ax$, remember to use $r = -ax$ in the geometric series.

Question 7

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

Explanation: This problem requires finding the power series representation of f(x) = 1/(1-3x) about x = 0. We start with the known geometric series formula 1/(1-u) = Σ(n=0 to ∞) u^n for |u| < 1. By substituting u = 3x, we get 1/(1-3x) = Σ(n=0 to ∞) (3x)^n = Σ(n=0 to ∞) 3^n x^n, which converges for |3x| < 1 or |x| < 1/3. Choice C might tempt students who think the series should start at n = 1, but the geometric series includes the n = 0 term, giving us 1 + 3x + 9x² + ... The key strategy is recognizing when a function matches the form 1/(1-u) and making the appropriate substitution.

Question 8

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

Explanation: This problem requires the power series for u(x) = e^(2x). The exponential series is e^y = Σ(n=0 to ∞) y^n/n! for all y. With y = 2x, we get u(x) = Σ(n=0 to ∞) (2x)^n/n! = Σ(n=0 to ∞) 2^n x^n/n!. Choice A incorrectly separates the 2 from x^n, giving only 2x^n instead of (2x)^n = 2^n x^n in the numerator. The key is to substitute the entire expression 2x into the exponential series, keeping 2x grouped together when raising to the nth power.

Question 9

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

Explanation: This question tests the skill of representing functions as power series, using the exponential series. The Maclaurin series for $e^y$ is $\sum_{n=0}^{\infty} \frac{y^n}{n!}$ for all real y. Substituting y = 2x gives $\sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}$, which is the series for e^{2x}. This holds for all real x since the exponential series converges everywhere. A tempting distractor is choice A, which has (n+1)! in the denominator, altering it to something like an integral of the exponential. A transferable strategy for representing functions as power series is to substitute scaled variables into known Taylor series like the exponential.

Question 10

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, applied in models like displacement with inverse trig functions. The series for arctan(x) comes from integrating the geometric series for 1/(1+t^2) = sum_{n=0}^∞ (-1)^n t^{2n} for |t|<1. Integrating from 0 to x yields arctan(x) = sum_{n=0}^∞ (-1)^n integral t^{2n} dt = sum_{n=0}^∞ (-1)^n x^{2n+1}/(2n+1). This is choice B. Choice D fails as a distractor by using (-1)^{n+1}, which negates the series and represents -arctan(x). A transferable strategy is to integrate known geometric series adaptations for inverse functions, carefully handling the powers and signs.

Question 11

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

Explanation: This problem requires finding the power series representation of h(x) = ln(1+x) about x = 0. We use the fact that ln(1+x) is the antiderivative of 1/(1+x), which has the series Σ(n=0 to ∞) (-1)^n x^n. Integrating term by term gives ∫[Σ(n=0 to ∞) (-1)^n x^n]dx = Σ(n=0 to ∞) (-1)^n x^(n+1)/(n+1), where the constant of integration is 0 since ln(1) = 0. This series converges for |x| < 1, matching the radius of convergence of 1/(1+x). Choice B might appeal to students who forget to adjust the denominator after integration, but integration increases the power by 1 and divides by the new power. Remember that integrating a power series term by term preserves the radius of convergence and requires adjusting both the power and coefficient.

Question 12

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

Explanation: This question tests the skill of representing functions as power series, particularly by adjusting the geometric series for different denominators. To find the series for $f(x) = \frac{1}{2-x}$, factor out the 2 from the denominator to get $\frac{1/2}{1 - x/2}$. This matches the geometric form $(1/2) \sum_{n=0}^{\infty} (x/2)^n$, which expands to $\sum_{n=0}^{\infty} \frac{x^n}{2^{n+1}}$ for $|x| < 2$. The adjustment ensures the common ratio is $x/2$ with $|x/2| < 1$. A tempting distractor is choice A, $\sum_{n=0}^{\infty} (x/2)^n$, which forgets to divide by 2 and represents $\frac{1}{1 - x/2} = \frac{2}{2-x}$ instead. A transferable strategy for representing functions as power series is to factor constants out of the denominator to align with the standard geometric series form.

Question 13

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

Explanation: This problem asks for the power series representation of v(x) = x/(2-x) about x = 0. We first rewrite v(x) = x · 1/(2-x) = x · (1/2) · 1/(1-x/2). Using the geometric series 1/(1-u) = Σ(n=0 to ∞) u^n with u = x/2 gives 1/(1-x/2) = Σ(n=0 to ∞) (x/2)^n. Therefore, v(x) = x · (1/2) · Σ(n=0 to ∞) (x/2)^n = (x/2) · Σ(n=0 to ∞) x^n/2^n = Σ(n=0 to ∞) x^(n+1)/2^(n+1), which converges for |x/2| < 1 or |x| < 2. Choice A incorrectly separates the factors of 2, placing 2^n in the denominator instead of 2^(n+1). When multiplying a series by x/k, both the power of x and the power of k increase by 1.

Question 14

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

Explanation: This problem asks for the power series representation of g(x) = 1/(4+x) about x = 0. We first factor out 4 from the denominator to get g(x) = 1/4 · 1/(1+x/4). Using the geometric series 1/(1-u) = Σ(n=0 to ∞) u^n with u = -x/4, we obtain 1/(1-(-x/4)) = Σ(n=0 to ∞) (-x/4)^n. Therefore, g(x) = 1/4 · Σ(n=0 to ∞) (-1)^n (x/4)^n = Σ(n=0 to ∞) (-1)^n x^n/4^(n+1), which converges for |x/4| < 1 or |x| < 4. Choice A incorrectly places the factor of 1/4 inside the sum without adjusting the exponent. When factoring constants from denominators, always factor completely before applying the geometric series formula.

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

Explanation: This problem requires finding the power series representation of s(x) = sin(3x) about x = 0. We use the known Maclaurin series sin(u) = Σ(n=0 to ∞) (-1)^n u^(2n+1)/(2n+1)! for all u. Substituting u = 3x gives sin(3x) = Σ(n=0 to ∞) (-1)^n (3x)^(2n+1)/(2n+1)!, which converges for all x. Choice E incorrectly separates 3^(2n+1) and x^(2n+1) but uses factorial (2n)! instead of (2n+1)! in the denominator, mixing up the sine and cosine series formulas. The key is maintaining the odd powers and odd factorials that characterize the sine series.

Question 16

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, applied in calibrations with binomial expansions. The generalized binomial series (1+y)^a = sum_{n=0}^∞ binom{a}{n} y^n for |y|<1. For u(x) = (1+x)^{1/2}, set a=1/2 and y=x, directly giving sum_{n=0}^∞ binom{1/2}{n} x^n. This is choice A. Choice D tempts with binom{-1/2}{n}, which represents 1/sqrt(1+x) instead. A transferable strategy is to use the binomial theorem for non-integer exponents, confirming the leading term matches the function at x=0.

Question 17

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, useful for modeling phenomena like temperature corrections. For h(x) = 1/(2+x), rewrite it as 1/(2(1 + x/2)) = (1/2) * 1/(1 - (-x/2)). Use the geometric series sum_{n=0}^∞ r^n = 1/(1-r) with r = -x/2, giving (1/2) sum_{n=0}^∞ (-x/2)^n = (1/2) sum_{n=0}^∞ (-1)^n x^n / 2^n = sum_{n=0}^∞ (-1)^n x^n / 2^{n+1}. This matches choice B. Choice A tempts by omitting the 1/2 factor, representing 1/(1 + x/2) instead, which is twice the actual function. A transferable strategy is to factor constants and adjust the geometric form to fit the denominator, ensuring the radius of convergence aligns.

Question 18

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, useful for response functions in systems. For v(x) = x^2 / (1-2x), first express 1/(1-2x) = sum_{n=0}^∞ (2x)^n = sum_{n=0}^∞ 2^n x^n for |2x|<1. Multiply by x^2: x^2 sum_{n=0}^∞ 2^n x^n = sum_{n=0}^∞ 2^n x^{n+2}. This matches choice C. Choice A distracts by lacking the x^2 factor, representing just 1/(1-2x). A transferable strategy is to multiply polynomial factors by geometric series, shifting indices to incorporate higher starting powers.

Question 19

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, crucial for growth models with higher powers. For s(x) = 1/(1-x)^2, start with the geometric series sum_{n=0}^∞ x^n = 1/(1-x), then differentiate to get sum_{n=1}^∞ n x^{n-1} = 1/(1-x)^2. Multiply by x: sum_{n=1}^∞ n x^n = x/(1-x)^2, but to match, note the standard form is sum_{n=0}^∞ (n+1) x^n = 1/(1-x)^2. This matches choice A. Choice B tempts by using n instead of n+1, representing the derivative form without adjustment. A transferable strategy is to differentiate known series and manipulate to achieve higher-order denominators, verifying with small n terms.

Question 20

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

Explanation: Representing functions as power series is a key skill in AP Calculus BC, vital for exponential functions in damping models. The exponential e^y = sum_{n=0}^∞ y^n / n! is well-known for all real y. Substitute y = 2x to get e^{2x} = sum_{n=0}^∞ (2x)^n / n! = sum_{n=0}^∞ 2^n x^n / n!. This corresponds to choice C. Choice D distracts by starting from n=1, omitting the constant term and representing e^{2x} - 1. A transferable strategy is to substitute linear arguments into fundamental series like exponential or trigonometric, preserving the factorial denominator.