Key Concepts and Formulas

1. Sum of the first $k$ natural numbers: The sum of an arithmetic progression $1 + 2 + \dots + k$ is given by the formula: $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$
2. Sum of the first $k$ cubes: The sum of the cubes of the first $k$ natural numbers is given by the formula: $\sum_{i=1}^{k} i^3 = \left( \frac{k(k+1)}{2} \right)^2$
3. Telescoping Series (Method of Differences): A series is called telescoping if most of the terms cancel out. This often happens when the general term $T_k$ can be expressed as a difference of two consecutive terms, i.e., $T_k = f(k) - f(k+1)$ or $T_k = f(k+1) - f(k)$. In such cases, the sum $S_n = \sum_{k=1}^{n} T_k$ simplifies significantly.

Step-by-Step Solution

Step 1: Identify the General Term ($T_k$) of the Series The given series is $S_n = \frac{1}{1^3} + \frac{1+2}{1^3+2^3} + \frac{1+2+3}{1^3+2^3+3^3} + \dots + \frac{1+2+\dots+n}{1^3+2^3+\dots+n^3}$. We need to find the general $k$-th term, $T_k$. Observing the pattern, the $k$-th term consists of the sum of the first $k$ natural numbers in the numerator and the sum of the cubes of the first $k$ natural numbers in the denominator. Therefore, the general term $T_k$ can be written as: $T_k = \frac{1+2+\dots+k}{1^3+2^3+\dots+k^3}$

Step 2: Simplify the General Term ($T_k$) using Standard Formulas We will use the formulas for the sum of the first $k$ natural numbers and the sum of the first $k$ cubes. The numerator is the sum of the first $k$ natural numbers: $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$ The denominator is the sum of the cubes of the first $k$ natural numbers: $\sum_{i=1}^{k} i^3 = \left( \frac{k(k+1)}{2} \right)^2$ Now, substitute these simplified expressions back into the formula for $T_k$: $T_k = \frac{\frac{k(k+1)}{2}}{\left( \frac{k(k+1)}{2} \right)^2}$ We can simplify this expression by canceling one power of the term in the denominator: $T_k = \frac{1}{\frac{k(k+1)}{2}}$ $T_k = \frac{2}{k(k+1)}$

Step 3: Express $T_k$ in a form suitable for Telescoping Series The term $T_k = \frac{2}{k(k+1)}$ can be decomposed using partial fractions into a difference of two terms, which is essential for a telescoping series. We can write $\frac{2}{k(k+1)}$ as $2 \left( \frac{1}{k(k+1)} \right)$. Using the standard partial fraction decomposition for $\frac{1}{k(k+1)}$: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$ Therefore, the general term $T_k$ becomes: $T_k = 2 \left( \frac{1}{k} - \frac{1}{k+1} \right)$

Step 4: Calculate the Sum ($S_n$) using the Telescoping Series Method Now we need to find the sum of the series $S_n = \sum_{k=1}^{n} T_k$. $S_n = \sum_{k=1}^{n} 2 \left( \frac{1}{k} - \frac{1}{k+1} \right)$ We can pull the constant factor 2 out of the summation: $S_n = 2 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right)$ Let's write out the terms of the sum to observe the cancellation: For $k=1$: $\frac{1}{1} - \frac{1}{2}$ For $k=2$: $\frac{1}{2} - \frac{1}{3}$ For $k=3$: $\frac{1}{3} - \frac{1}{4}$ ... For $k=n$: $\frac{1}{n} - \frac{1}{n+1}$

Summing these terms, we see that the intermediate terms cancel out (e.g., $-\frac{1}{2}$ cancels with $+\frac{1}{2}$, $-\frac{1}{3}$ cancels with $+\frac{1}{3}$, and so on). This is a telescoping sum. $\sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$ $= 1 - \frac{1}{n+1}$ Now, substitute this back into the expression for $S_n$: $S_n = 2 \left( 1 - \frac{1}{n+1} \right)$ Simplify the expression inside the parentheses: $S_n = 2 \left( \frac{(n+1)-1}{n+1} \right)$ $S_n = 2 \left( \frac{n}{n+1} \right)$ $S_n = \frac{2n}{n+1}$

Step 5: Solve for $n$ using the given condition The problem states that $100 S_n = n$. Substitute the expression for $S_n$ we found: $100 \left( \frac{2n}{n+1} \right) = n$ $\frac{200n}{n+1} = n$ Since $n$ represents the number of terms in a series, $n$ must be a positive integer ($n \ge 1$). Thus, $n \ne 0$, and we can divide both sides by $n$: $\frac{200}{n+1} = 1$ Now, multiply both sides by $(n+1)$: $200 = n+1$ Subtract 1 from both sides to solve for $n$: $n = 200 - 1$ $n = 199$

Common Mistakes & Tips

• Incorrectly applying summation formulas: Ensure you are using the correct formulas for the sum of natural numbers and the sum of cubes.
• Algebraic errors during simplification: Be meticulous when simplifying fractions, especially complex fractions like $T_k$.
• Failure to recognize telescoping series: The form $\frac{1}{k(k+1)}$ is a strong indicator of a telescoping series, which can be efficiently handled using partial fraction decomposition.
• Division by zero: When solving the final equation, remember that $n$ must be a positive integer, so dividing by $n$ is permissible.

Summary

The problem involves calculating the sum of a series where the $k$-th term is the ratio of the sum of the first $k$ natural numbers to the sum of the cubes of the first $k$ natural numbers. By using the standard formulas for these sums, we simplified the general term $T_k$ to $\frac{2}{k(k+1)}$. This form allowed us to express $T_k$ as a difference of consecutive terms using partial fractions, transforming the series into a telescoping sum. Evaluating the telescoping sum yielded $S_n = \frac{2n}{n+1}$. Finally, using the given condition $100 S_n = n$, we solved for $n$ and found it to be 199.

The final answer is $\boxed{199}$.