Key Concepts and Formulas

• Telescoping Series: A series where intermediate terms cancel out, simplifying the sum to a few remaining terms. This often arises from expressing a term as a difference of two consecutive terms in a sequence.
• Partial Fraction Decomposition (Implicit): The technique used to rewrite a fraction into a sum or difference of simpler fractions. For terms of the form $\frac{1}{ab}$, we often aim for $\frac{1}{b-a}\left(\frac{1}{a} - \frac{1}{b}\right)$.

Step-by-Step Solution

Step 1: Identify the Structure of the General Term and Apply Telescoping Technique

The given series is: $S = \frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$ The general term of this series can be represented as: $T_k = \frac{1}{(1+(k-1)d)(1+kd)}$ where $k$ ranges from $1$ to $10$.

To apply the telescoping series concept, we need to express each term $T_k$ as a difference of two terms. Observe the denominator: the two factors differ by $(1+kd) - (1+(k-1)d) = d$. We can rewrite $T_k$ by multiplying and dividing by $d$: $T_k = \frac{1}{d} \cdot \frac{d}{(1+(k-1)d)(1+kd)}$ Now, we use the difference in the numerator: $T_k = \frac{1}{d} \left( \frac{(1+kd) - (1+(k-1)d)}{(1+(k-1)d)(1+kd)} \right)$ $T_k = \frac{1}{d} \left( \frac{1}{1+(k-1)d} - \frac{1}{1+kd} \right)$

Step 2: Sum the Series Using the Telescoping Property

Now, we sum these terms from $k=1$ to $k=10$: $S = \sum_{k=1}^{10} T_k = \sum_{k=1}^{10} \frac{1}{d} \left( \frac{1}{1+(k-1)d} - \frac{1}{1+kd} \right)$ We can pull the constant factor $\frac{1}{d}$ out of the summation: $S = \frac{1}{d} \sum_{k=1}^{10} \left( \frac{1}{1+(k-1)d} - \frac{1}{1+kd} \right)$ Let's write out the first few terms and the last few terms to see the cancellation: For $k=1$: $\left( \frac{1}{1+0d} - \frac{1}{1+1d} \right) = \left( \frac{1}{1} - \frac{1}{1+d} \right)$ For $k=2$: $\left( \frac{1}{1+1d} - \frac{1}{1+2d} \right)$ For $k=3$: $\left( \frac{1}{1+2d} - \frac{1}{1+3d} \right)$ ... For $k=9$: $\left( \frac{1}{1+8d} - \frac{1}{1+9d} \right)$ For $k=10$: $\left( \frac{1}{1+9d} - \frac{1}{1+10d} \right)$

When we add these terms, the intermediate terms cancel out: $-\frac{1}{1+d}$ cancels with $+\frac{1}{1+d}$, $-\frac{1}{1+2d}$ cancels with $+\frac{1}{1+2d}$, and so on, until $-\frac{1}{1+9d}$ cancels with $+\frac{1}{1+9d}$. The sum simplifies to: $S = \frac{1}{d} \left( \frac{1}{1} - \frac{1}{1+10d} \right)$

Step 3: Solve for d Using the Given Sum

We are given that the sum of the series is equal to 5: $S = 5$ Substituting our simplified expression for $S$: $5 = \frac{1}{d} \left( 1 - \frac{1}{1+10d} \right)$ First, simplify the expression inside the parentheses: $1 - \frac{1}{1+10d} = \frac{(1+10d) - 1}{1+10d} = \frac{10d}{1+10d}$ Now substitute this back into the equation: $5 = \frac{1}{d} \left( \frac{10d}{1+10d} \right)$ The $d$ in the numerator and the $d$ in the denominator cancel out: $5 = \frac{10}{1+10d}$ Now, we solve for $d$: $5(1+10d) = 10$ Divide both sides by 5: $1+10d = 2$ Subtract 1 from both sides: $10d = 1$ Divide by 10: $d = \frac{1}{10}$

Step 4: Calculate the Value of 50d

The question asks for the value of $50d$. We found that $d = \frac{1}{10}$. $50d = 50 \times \frac{1}{10}$ $50d = \frac{50}{10}$ $50d = 5$

Common Mistakes & Tips

• Forgetting to factor out $\frac{1}{d}$: The telescoping form relies on expressing each term as $\frac{1}{d} \left( \frac{1}{A} - \frac{1}{B} \right)$. Failing to include the $\frac{1}{d}$ factor will lead to an incorrect value for $d$.
• Algebraic errors in simplification: Be meticulous when combining fractions or solving the final equation for $d$. A small arithmetic error can lead to the wrong answer.
• Incorrectly identifying the last term: Ensure the summation is performed for the correct number of terms (10 terms in this case), and that the last term in the telescoping sum is correctly identified as $-\frac{1}{1+10d}$.

Summary

This problem is a classic example of a telescoping series. By recognizing that each term in the series can be expressed as a difference of two fractions, we were able to simplify the entire sum to just two terms. This allowed us to set up an equation with the given sum and solve for the unknown value of $d$. Finally, we used the obtained value of $d$ to calculate the required expression $50d$.

The final answer is \boxed{5}.