Key Concepts and Formulas

• Partial Fraction Decomposition: A method to express a rational function as a sum of simpler rational functions. For a term of the form $\frac{1}{n(n+a)(n+b)}$, we can decompose it into $\frac{A}{n} + \frac{B}{n+a} + \frac{C}{n+b}$.
• Telescoping Series: A series where most of the terms cancel out when the sum is calculated. This typically occurs when the general term can be expressed as a difference of consecutive terms of a sequence.
• Summation Notation: The use of $\sum$ to represent a sum of a sequence of terms.

Step-by-Step Solution

Step 1: Decomposing the General Term using Partial Fractions The general term of the series is $\frac{1}{n(n+1)(n+2)}$. We aim to decompose this into simpler fractions using partial fractions. We assume: $\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$ Multiplying both sides by $n(n+1)(n+2)$ gives: $1 = A(n+1)(n+2) + B n(n+2) + C n(n+1)$ To find the coefficients $A$, $B$, and $C$, we can substitute strategic values of $n$:

• Let $n=0$: $1 = A(1)(2) \implies 2A = 1 \implies A = \frac{1}{2}$.
• Let $n=-1$: $1 = B(-1)(-1+2) \implies 1 = B(-1)(1) \implies -B = 1 \implies B = -1$.
• Let $n=-2$: $1 = C(-2)(-2+1) \implies 1 = C(-2)(-1) \implies 2C = 1 \implies C = \frac{1}{2}$. So, the decomposition is: $\frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} = \frac{1}{2} \left( \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \right)$ We can further rewrite the expression inside the parenthesis to facilitate telescoping: $\frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} = \left(\frac{1}{n} - \frac{1}{n+1}\right) - \left(\frac{1}{n+1} - \frac{1}{n+2}\right)$ This simplifies to: $\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}$ Therefore, the general term can be written as: $\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$

Step 2: Applying the Telescoping Series to the Given Sum The given sum is $S = \frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}$. This can be written in summation notation as $S = \sum_{n=2}^{100} \frac{1}{n(n+1)(n+2)}$. Using the decomposition from Step 1: $S = \sum_{n=2}^{100} \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$ $S = \frac{1}{2} \sum_{n=2}^{100} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$ This is a telescoping series. Let's write out the terms: For $n=2$: $\frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4}$ For $n=3$: $\frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5}$ For $n=4$: $\frac{1}{4 \cdot 5} - \frac{1}{5 \cdot 6}$ ... For $n=100$: $\frac{1}{100 \cdot 101} - \frac{1}{101 \cdot 102}$ When we sum these terms, the intermediate terms cancel out: $S = \frac{1}{2} \left[ \left(\frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4}\right) + \left(\frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5}\right) + \dots + \left(\frac{1}{100 \cdot 101} - \frac{1}{101 \cdot 102}\right) \right]$ $S = \frac{1}{2} \left[ \frac{1}{2 \cdot 3} - \frac{1}{101 \cdot 102} \right]$

Step 3: Simplifying the Sum Now, we simplify the expression for $S$: $S = \frac{1}{2} \left[ \frac{1}{6} - \frac{1}{10302} \right]$ To subtract the fractions, we find a common denominator: $S = \frac{1}{2} \left[ \frac{10302 - 6}{6 \times 10302} \right] = \frac{1}{2} \left[ \frac{10296}{61812} \right]$ We can simplify the fraction $\frac{10296}{61812}$: Divide by 6: $\frac{1716}{10302}$ Divide by 2: $\frac{858}{5151}$ Divide by 3: $\frac{286}{1717}$ So, $S = \frac{1}{2} \times \frac{286}{1717} = \frac{143}{1717}$. We notice that $1717 = 17 \times 101$. Therefore, $S = \frac{143}{17 \times 101}$.

Step 4: Finding the value of k We are given that $S = \frac{k}{101}$. Equating our result with the given form: $\frac{k}{101} = \frac{143}{17 \times 101}$ Multiplying both sides by 101, we get: $k = \frac{143}{17}$

Step 5: Calculating 34k We need to find the value of $34k$. $34k = 34 \times \frac{143}{17}$ $34k = (2 \times 17) \times \frac{143}{17}$ $34k = 2 \times 143$ $34k = 286$

Common Mistakes & Tips

• Incorrect Partial Fraction Coefficients: Carefully check the values of A, B, and C by substituting values of $n$ or by comparing coefficients of powers of $n$.
• Error in Telescoping Cancellation: Write out the first few and last few terms of the series explicitly to ensure correct identification of the terms that remain.
• Arithmetic Errors: Double-check all calculations, especially when simplifying fractions and performing multiplications.
• Final Answer Misinterpretation: Ensure you are answering the question asked (e.g., $34k$) and not just an intermediate value like $k$.

Summary The problem is solved by first decomposing the general term of the series, $\frac{1}{n(n+1)(n+2)}$, into simpler partial fractions. This decomposition allows us to express the general term as a difference of two simpler fractions, $\frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$. When this form is summed from $n=2$ to $n=100$, it results in a telescoping series where most terms cancel out, leaving only the first and last terms. After simplifying the resulting expression and equating it to the given form $\frac{k}{101}$, we find the value of $k$ and then compute $34k$.

The final answer is $\boxed{286}$.