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 $x = a$.

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 $x$.

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.

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 $q(x) = \arctan(x)$ 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 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.

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 $x^2$.

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 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.

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.