Key Concepts and Formulas

• Sum of the first $n$ natural numbers: $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$.
• Partial Fraction Decomposition: Expressing a rational function as a sum or difference of simpler rational functions. For $\frac{1}{n(n+1)}$, we can write it as $\frac{1}{n} - \frac{1}{n+1}$.
• Telescoping Series: A series where most of the intermediate terms cancel out when summed, typically of the form $\sum_{n=1}^{N} (f(n) - f(n+1))$ or $\sum_{n=1}^{N} (f(n) - f(n+k))$.

Step-by-Step Solution

Step 1: Identify the General Term ($T_n$) of the Series The given series is $S = 1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \dots + {1 \over {1 + 2 + 3 + \dots + 11}}$. Let's denote the $n$-th term of the series as $T_n$. The denominator of each term is the sum of the first $n$ natural numbers. Using the formula for the sum of the first $n$ natural numbers, the denominator of the $n$-th term is $\frac{n(n+1)}{2}$. Therefore, the general term $T_n$ is: $T_n = {1 \over {1 + 2 + 3 + \dots + n}} = {1 \over {\frac{n(n+1)}{2}}}$ Simplifying this, we get: $T_n = {2 \over {n(n+1)}}$ We can verify this for the first term ($n=1$): $T_1 = \frac{2}{1(1+1)} = \frac{2}{2} = 1$, which matches the first term of the series.

Step 2: Decompose the General Term using Partial Fractions To apply the telescoping sum method, we need to express $T_n$ as a difference of two terms. We can rewrite $T_n$ as: $T_n = 2 \times {1 \over {n(n+1)}}$ Now, we decompose $\frac{1}{n(n+1)}$ using partial fractions. We know that: ${1 \over {n(n+1)}} = {1 \over n} - {1 \over {n+1}}$ Substituting this back into the expression for $T_n$: $T_n = 2 \left( {1 \over n} - {1 \over {n+1}} \right)$ This form is crucial for the telescoping sum.

Step 3: Calculate the Sum of the Series using the Telescoping Method The given series is the sum of terms from $n=1$ to $n=11$. Let $S_{11}$ be the sum of these terms. $S_{11} = \sum_{n=1}^{11} T_n = \sum_{n=1}^{11} 2 \left( {1 \over n} - {1 \over {n+1}} \right)$ We can factor out the constant 2: $S_{11} = 2 \sum_{n=1}^{11} \left( {1 \over n} - {1 \over {n+1}} \right)$ Now, let's write out the terms of the summation: For $n=1$: ${1 \over 1} - {1 \over 2}$ For $n=2$: ${1 \over 2} - {1 \over 3}$ For $n=3$: ${1 \over 3} - {1 \over 4}$ ... For $n=10$: ${1 \over 10} - {1 \over 11}$ For $n=11$: ${1 \over 11} - {1 \over 12}$

When we sum these terms, the intermediate terms cancel out: $S_{11} = 2 \left[ \left( {1 \over 1} - {1 \over 2} \right) + \left( {1 \over 2} - {1 \over 3} \right) + \left( {1 \over 3} - {1 \over 4} \right) + \dots + \left( {1 \over 10} - {1 \over 11} \right) + \left( {1 \over 11} - {1 \over 12} \right) \right]$ The $-{1 \over 2}$ cancels with $+{1 \over 2}$, $-{1 \over 3}$ cancels with $+{1 \over 3}$, and so on, until $-{1 \over 11}$ cancels with $+{1 \over 11}$. This leaves us with only the first part of the first term and the last part of the last term: $S_{11} = 2 \left[ {1 \over 1} - {1 \over 12} \right]$

Step 4: Simplify the Result Now, we simplify the expression inside the brackets and then multiply by 2: $S_{11} = 2 \left[ {{12 - 1} \over 12} \right]$ $S_{11} = 2 \left[ {{11} \over 12} \right]$ $S_{11} = {{2 \times 11} \over 12}$ $S_{11} = {{22} \over 12}$ Finally, simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2: $S_{11} = {{11} \over 6}$

Common Mistakes & Tips

• Incorrect General Term: Ensure the formula for the general term correctly represents each term in the series, especially the first term.
• Forgetting the Constant Factor: The constant '2' in $T_n = \frac{2}{n(n+1)}$ must be carried through the entire summation process.
• Off-by-One Error in Limits: The series goes up to the term involving $1+2+\dots+11$, which means there are 11 terms in total (from $n=1$ to $n=11$).

Summary

The problem requires summing a series whose terms are reciprocals of the sums of consecutive natural numbers. By first finding the general term $T_n = \frac{2}{n(n+1)}$ and then decomposing it into partial fractions as $T_n = 2 \left( \frac{1}{n} - \frac{1}{n+1} \right)$, we can apply the telescoping sum method. This method leads to the cancellation of most intermediate terms, leaving a simple expression. Summing the 11 terms of the series results in $\frac{11}{6}$.

The final answer is $\boxed{\frac{11}{6}}$ which corresponds to option (B).