Key Concepts and Formulas

• Trigonometric Identity for Arctangent: The key identity used here is $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. This allows us to express a difference of arctangents as a single arctangent.
• Telescoping Series: A series where most of the terms cancel out. For a sum of the form $\sum_{r=1}^k (f(r+1) - f(r))$, the sum simplifies to $f(k+1) - f(1)$.
• Limits of Arctangent: The limit of $\tan^{-1}(x)$ as $x \to \infty$ is $\frac{\pi}{2}$, and the limit as $x \to -\infty$ is $-\frac{\pi}{2}$.
• Complementary Angle Identity: $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$. This is used to convert an arctangent to a cotangent.

Step-by-Step Solution

Step 1: Rewrite the general term of the series The given general term is $a_r = \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$. We want to manipulate this to fit the form $\tan^{-1}(x) - \tan^{-1}(y)$. $a_r = \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$ Divide the numerator and denominator by $2^{2r+1}$: $a_r = \tan^{-1}\left(\frac{\frac{6^r}{2^{2r+1}}}{\frac{2^{2r+1}}{2^{2r+1}} + \frac{3^{2r+1}}{2^{2r+1}}}\right) = \tan^{-1}\left(\frac{\frac{(2 \cdot 3)^r}{2^{2r+1}}}{1 + \left(\frac{3}{2}\right)^{2r+1}}\right)$ Simplify the numerator: $\frac{(2 \cdot 3)^r}{2^{2r+1}} = \frac{2^r \cdot 3^r}{2^{2r} \cdot 2^1} = \frac{3^r}{2^{r+1}}$ So, the term becomes: $a_r = \tan^{-1}\left(\frac{\frac{3^r}{2^{r+1}}}{1 + \left(\frac{3}{2}\right)^{2r+1}}\right)$ This doesn't immediately look like the required form. Let's try dividing by $3^{2r+1}$ instead. $a_r = \tan^{-1}\left(\frac{\frac{6^r}{3^{2r+1}}}{\frac{2^{2r+1}}{3^{2r+1}} + \frac{3^{2r+1}}{3^{2r+1}}}\right) = \tan^{-1}\left(\frac{\frac{(2 \cdot 3)^r}{3^{2r+1}}}{\left(\frac{2}{3}\right)^{2r+1} + 1}\right)$ Simplify the numerator: $\frac{(2 \cdot 3)^r}{3^{2r+1}} = \frac{2^r \cdot 3^r}{3^{2r} \cdot 3^1} = \frac{2^r}{3^{r+1}}$ So, the term becomes: $a_r = \tan^{-1}\left(\frac{\frac{2^r}{3^{r+1}}}{1 + \left(\frac{2}{3}\right)^{2r+1}}\right)$ This still doesn't match the pattern. Let's go back to the original expression and try to manipulate it differently. The goal is to get it into the form $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. Let's consider the denominator $2^{2r+1} + 3^{2r+1} = 2 \cdot 4^r + 3 \cdot 9^r$. The term is $\tan^{-1}\left(\frac{6^r}{2 \cdot 4^r + 3 \cdot 9^r}\right)$. Divide numerator and denominator by $3^{2r+1}$: $a_r = \tan^{-1}\left(\frac{\frac{6^r}{3^{2r+1}}}{\frac{2^{2r+1}}{3^{2r+1}} + 1}\right) = \tan^{-1}\left(\frac{\frac{2^r 3^r}{3^{2r} 3}}{(\frac{2}{3})^{2r+1} + 1}\right) = \tan^{-1}\left(\frac{\frac{2^r}{3^{r+1}}}{1 + (\frac{2}{3})^{2r+1}}\right)$ This is still not matching. Let's try dividing by $2^{2r+1}$ again, but rewrite the numerator $6^r$ as $3^r \cdot 2^r$. $a_r = \tan^{-1}\left(\frac{2^r \cdot 3^r}{2^{2r+1} + 3^{2r+1}}\right)$ Divide numerator and denominator by $2^{2r+1}$: $a_r = \tan^{-1}\left(\frac{\frac{2^r \cdot 3^r}{2^{2r+1}}}{1 + \frac{3^{2r+1}}{2^{2r+1}}}\right) = \tan^{-1}\left(\frac{\frac{3^r}{2^{r+1}}}{1 + \left(\frac{3}{2}\right)^{2r+1}}\right)$ Let's try to represent the numerator $6^r$ as a difference of terms that will create the $1+xy$ in the denominator. Consider the form $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$. We have $\tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$. Let's try to make the denominator $1 + (\frac{3}{2})^{2r+1} (\frac{2}{3})^{2r+1} = 1+1$. This is not helpful.

Let's look at the structure of the options. They involve $\tan^{-1}(3/2)$ or $\cot^{-1}(3/2)$. This suggests that terms involving $(3/2)^r$ are likely to appear.

Let's rewrite the general term as: $a_r = \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right) = \tan^{-1}\left(\frac{2^r 3^r}{2 \cdot 4^r + 3 \cdot 9^r}\right)$ Divide numerator and denominator by $3^{2r+1}$: $a_r = \tan^{-1}\left(\frac{\frac{2^r 3^r}{3^{2r+1}}}{\frac{2^{2r+1}}{3^{2r+1}} + 1}\right) = \tan^{-1}\left(\frac{\frac{2^r}{3^{r+1}}}{(\frac{2}{3})^{2r+1} + 1}\right)$ This still doesn't look right. Let's try dividing the numerator and denominator by $2^{2r+1}$: $a_r = \tan^{-1}\left(\frac{\frac{6^r}{2^{2r+1}}}{1 + \frac{3^{2r+1}}{2^{2r+1}}}\right) = \tan^{-1}\left(\frac{\frac{2^r 3^r}{2^{2r+1}}}{1 + (\frac{3}{2})^{2r+1}}\right) = \tan^{-1}\left(\frac{\frac{3^r}{2^{r+1}}}{1 + (\frac{3}{2})^{2r+1}}\right)$ Consider the term $\frac{3^r}{2^{r+1}}$. We want to express this in a way that fits the $\tan^{-1}(x) - \tan^{-1}(y)$ formula. Let's try to manipulate the numerator $6^r$ as a difference. The denominator is $2^{2r+1} + 3^{2r+1}$. Consider the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. We have $\frac{6^r}{2^{2r+1} + 3^{2r+1}}$. Let's try to make the denominator look like $1+xy$. Divide by $3^{2r+1}$: $\tan^{-1}\left(\frac{\frac{6^r}{3^{2r+1}}}{1 + \frac{2^{2r+1}}{3^{2r+1}}}\right) = \tan^{-1}\left(\frac{\frac{2^r 3^r}{3^{2r} 3}}{1 + (\frac{2}{3})^{2r+1}}\right) = \tan^{-1}\left(\frac{\frac{2^r}{3^{r+1}}}{1 + (\frac{2}{3})^{2r+1}}\right)$ This is not working as expected. Let's re-examine the given solution's first step: $S_k = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$ This step is incorrect. The original term is $\frac{6^r}{2^{2r+1} + 3^{2r+1}}$. Let's try to directly obtain the form $\tan^{-1}(A) - \tan^{-1}(B)$ from the original expression. $\frac{6^r}{2^{2r+1} + 3^{2r+1}} = \frac{2^r 3^r}{2 \cdot 4^r + 3 \cdot 9^r}$ Divide numerator and denominator by $3^{2r+1}$: $\frac{\frac{2^r 3^r}{3^{2r+1}}}{1 + \frac{2^{2r+1}}{3^{2r+1}}} = \frac{\frac{2^r}{3^{r+1}}}{1 + (\frac{2}{3})^{2r+1}}$ This still doesn't lead to a simple difference.

Let's try dividing by $2^{2r+1}$ instead of $3^{2r+1}$: $\frac{\frac{6^r}{2^{2r+1}}}{1 + \frac{3^{2r+1}}{2^{2r+1}}} = \frac{\frac{2^r 3^r}{2^{2r+1}}}{1 + (\frac{3}{2})^{2r+1}} = \frac{\frac{3^r}{2^{r+1}}}{1 + (\frac{3}{2})^{2r+1}}$ This also doesn't directly yield the form $\frac{x-y}{1+xy}$.

Let's consider the form $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$. We have $\tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$. Let's try to force the denominator into the form $1+xy$. Divide by $3^{2r+1}$: $\tan^{-1}\left(\frac{\frac{6^r}{3^{2r+1}}}{1 + \frac{2^{2r+1}}{3^{2r+1}}}\right) = \tan^{-1}\left(\frac{\frac{2^r}{3^{r+1}}}{1 + (\frac{2}{3})^{2r+1}}\right)$ This is not leading to the desired form.

Let's re-examine the original problem and the provided solution's intermediate steps carefully. The key is to express the argument of the arctan in the form $\frac{x-y}{1+xy}$. The argument is $\frac{6^r}{2^{2r+1} + 3^{2r+1}}$. Let's divide the numerator and denominator by $3^{2r+1}$: $\frac{\frac{6^r}{3^{2r+1}}}{1 + \frac{2^{2r+1}}{3^{2r+1}}} = \frac{\frac{2^r 3^r}{3^{2r} 3}}{1 + (\frac{2}{3})^{2r+1}} = \frac{\frac{2^r}{3^{r+1}}}{1 + (\frac{2}{3})^{2r+1}}$ This is still not in the form $\frac{x-y}{1+xy}$.

Let's try dividing the numerator and denominator by $2^{2r+1}$: $\frac{\frac{6^r}{2^{2r+1}}}{1 + \frac{3^{2r+1}}{2^{2r+1}}} = \frac{\frac{2^r 3^r}{2^{2r} 2}}{1 + (\frac{3}{2})^{2r+1}} = \frac{\frac{3^r}{2^{r+1}}}{1 + (\frac{3}{2})^{2r+1}}$ This also does not directly match.

Let's consider the target form $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$. We have $\frac{6^r}{2^{2r+1} + 3^{2r+1}}$. Let's try to make the denominator $1+AB$. Divide numerator and denominator by $3^{2r+1}$: $\frac{\frac{6^r}{3^{2r+1}}}{1 + \frac{2^{2r+1}}{3^{2r+1}}} = \frac{\frac{2^r 3^r}{3^{2r} 3}}{1 + (\frac{2}{3})^{2r+1}} = \frac{\frac{1}{3} (\frac{2}{3})^r}{1 + (\frac{2}{3})^{2r+1}}$ This is not working.

Let's divide the numerator and denominator by $2^{2r+1}$: $\frac{\frac{6^r}{2^{2r+1}}}{1 + \frac{3^{2r+1}}{2^{2r+1}}} = \frac{\frac{2^r 3^r}{2^{2r} 2}}{1 + (\frac{3}{2})^{2r+1}} = \frac{\frac{1}{2} (\frac{3}{2})^r}{1 + (\frac{3}{2})^{2r+1}}$ This is still not in the form.

Let's try to manipulate the expression to get terms like $(\frac{3}{2})^{r+1}$ and $(\frac{3}{2})^r$. Consider the term $\tan^{-1}\left(\frac{\frac{3^r}{2^{r+1}}}{1 + (\frac{3}{2})^{2r+1}}\right)$. Let's try to use the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. We need $x-y = \frac{3^r}{2^{r+1}}$ and $1+xy = 1 + (\frac{3}{2})^{2r+1}$. This implies $xy = (\frac{3}{2})^{2r+1}$. If $x = (\frac{3}{2})^{r+1}$ and $y = (\frac{3}{2})^r$, then $xy = (\frac{3}{2})^{2r+1}$. Then $x-y = (\frac{3}{2})^{r+1} - (\frac{3}{2})^r = (\frac{3}{2})^r (\frac{3}{2} - 1) = (\frac{3}{2})^r (\frac{1}{2}) = \frac{3^r}{2^r} \frac{1}{2} = \frac{3^r}{2^{r+1}}$. This matches the numerator we found earlier!

So, the general term is: $a_r = \tan^{-1}\left(\frac{\frac{3^r}{2^{r+1}}}{1 + (\frac{3}{2})^{2r+1}}\right)$ We can rewrite this as: $a_r = \tan^{-1}\left(\frac{(\frac{3}{2})^{r+1} - (\frac{3}{2})^r}{1 + (\frac{3}{2})^{r+1} (\frac{3}{2})^r}\right)$ This is because $(\frac{3}{2})^{r+1} - (\frac{3}{2})^r = (\frac{3}{2})^r (\frac{3}{2} - 1) = (\frac{3}{2})^r (\frac{1}{2}) = \frac{3^r}{2^r} \frac{1}{2} = \frac{3^r}{2^{r+1}}$. And $1 + (\frac{3}{2})^{r+1} (\frac{3}{2})^r = 1 + (\frac{3}{2})^{2r+1}$. Thus, by the arctangent subtraction formula, $a_r = \tan^{-1}\left(\left(\frac{3}{2}\right)^{r+1}\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^r\right)$

Step 2: Evaluate the sum $S_k$ Now we have $S_k = \sum_{r=1}^k \left[\tan^{-1}\left(\left(\frac{3}{2}\right)^{r+1}\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^r\right)\right]$. This is a telescoping series. Let $f(r) = \tan^{-1}\left(\left(\frac{3}{2}\right)^r\right)$. Then the sum is $\sum_{r=1}^k [f(r+1) - f(r)]$. The sum expands as: $S_k = \left[\tan^{-1}\left(\left(\frac{3}{2}\right)^2\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^1\right)\right]$ $+ \left[\tan^{-1}\left(\left(\frac{3}{2}\right)^3\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^2\right)\right]$ $+ \left[\tan^{-1}\left(\left(\frac{3}{2}\right)^4\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^3\right)\right]$ $+ \dots$ $+ \left[\tan^{-1}\left(\left(\frac{3}{2}\right)^{k+1}\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^k\right)\right]$ All intermediate terms cancel out. The sum simplifies to: $S_k = \tan^{-1}\left(\left(\frac{3}{2}\right)^{k+1}\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^1\right)$

Step 3: Evaluate the limit as $k \to \infty$ We need to find $\lim_{k \to \infty} S_k$. $\lim_{k \to \infty} S_k = \lim_{k \to \infty} \left[\tan^{-1}\left(\left(\frac{3}{2}\right)^{k+1}\right) - \tan^{-1}\left(\frac{3}{2}\right)\right]$ As $k \to \infty$, the term $\left(\frac{3}{2}\right)^{k+1}$ approaches infinity because $\frac{3}{2} > 1$. We know that $\lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2}$. Therefore, $\lim_{k \to \infty} S_k = \frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right)$

Step 4: Simplify the result using complementary angle identity We use the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$, which can be rearranged as $\frac{\pi}{2} - \tan^{-1}(x) = \cot^{-1}(x)$. So, $\frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right) = \cot^{-1}\left(\frac{3}{2}\right)$

Common Mistakes & Tips

• Incorrect manipulation of the general term: The most common mistake is not being able to express the argument of the arctan in the form $\frac{x-y}{1+xy}$. Careful algebraic manipulation is crucial.
• Errors in telescoping sum cancellation: Ensure that the terms are correctly identified as $f(r+1) - f(r)$ and that the cancellation is done properly, leaving only the first part of the first term and the second part of the last term.
• Misunderstanding limits of arctan: Remember that $\lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \tan^{-1}(x) = -\frac{\pi}{2}$.

Summary

The problem involves finding the sum of a series of arctangent terms and then evaluating its limit. The key step is to rewrite the general term of the series, $\tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$, into the form $\tan^{-1}(A) - \tan^{-1}(B)$ by recognizing that $A = (\frac{3}{2})^{r+1}$ and $B = (\frac{3}{2})^r$. This allows the series to become a telescoping sum, simplifying to $\tan^{-1}\left(\left(\frac{3}{2}\right)^{k+1}\right) - \tan^{-1}\left(\frac{3}{2}\right)$. Taking the limit as $k \to \infty$ leads to $\frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right)$, which by the complementary angle identity is equal to $\cot^{-1}\left(\frac{3}{2}\right)$.

The final answer is \boxed{{\cot ^{ - 1}}\left( {{3 \over 2}} \right)}. This corresponds to option (A).