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 $x = a$.

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.

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.

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.

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.

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 $a$.

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.

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.

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.

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 $x$.