Key Concepts and Formulas

• Finding the $n$-th term from the sum of the first $n$ terms: If $S_n = a_1 + a_2 + \dots + a_n$, then the $n$-th term $a_n$ can be found using $a_n = S_n - S_{n-1}$ for $n \ge 2$, and $a_1 = S_1$.
• Summation of products of consecutive integers: The sum of products of three consecutive integers can be simplified using the identity $\sum_{k=1}^{n} k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$.
• Prime Factorization: The process of expressing a composite number as a product of its prime factors.

Step-by-Step Solution

Step 1: Find the general term $a_n$. We are given the sum of the first $n$ terms as $S_n = a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$. To find the $n$-th term, $a_n$, we use the relationship $a_n = S_n - S_{n-1}$ for $n \ge 2$. First, let's find $S_{n-1}$: $S_{n-1} = \frac{(n-1)^2 + 3(n-1)}{((n-1)+1)((n-1)+2)} = \frac{n^2 - 2n + 1 + 3n - 3}{n(n+1)} = \frac{n^2 + n - 2}{n(n+1)}$ Now, we calculate $a_n$: $a_n = S_n - S_{n-1} = \frac{n^2 + 3n}{(n+1)(n+2)} - \frac{n^2 + n - 2}{n(n+1)}$ To subtract these fractions, we find a common denominator, which is $n(n+1)(n+2)$: $a_n = \frac{n(n^2 + 3n) - (n+2)(n^2 + n - 2)}{n(n+1)(n+2)}$ $a_n = \frac{(n^3 + 3n^2) - (n^3 + n^2 - 2n + 2n^2 + 2n - 4)}{n(n+1)(n+2)}$ $a_n = \frac{n^3 + 3n^2 - (n^3 + 3n^2 - 4)}{n(n+1)(n+2)}$ $a_n = \frac{4}{n(n+1)(n+2)}$ This formula is valid for $n \ge 2$. Let's check for $n=1$. $S_1 = \frac{1^2 + 3(1)}{(1+1)(1+2)} = \frac{4}{2 \cdot 3} = \frac{2}{3}$. Using the derived formula for $a_n$ with $n=1$: $a_1 = \frac{4}{1(1+1)(1+2)} = \frac{4}{1 \cdot 2 \cdot 3} = \frac{4}{6} = \frac{2}{3}$. Since $a_1 = S_1$, the formula $a_n = \frac{4}{n(n+1)(n+2)}$ is valid for all $n \ge 1$.

Step 2: Calculate $\sum_{k=1}^{10} \frac{1}{a_k}$. From Step 1, we have $a_k = \frac{4}{k(k+1)(k+2)}$. Therefore, $\frac{1}{a_k} = \frac{k(k+1)(k+2)}{4}$. We need to calculate: $\sum_{k=1}^{10} \frac{1}{a_k} = \sum_{k=1}^{10} \frac{k(k+1)(k+2)}{4} = \frac{1}{4} \sum_{k=1}^{10} k(k+1)(k+2)$ We use the formula for the sum of products of consecutive integers: $\sum_{k=1}^{n} k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$. For $n=10$: $\sum_{k=1}^{10} k(k+1)(k+2) = \frac{10(10+1)(10+2)(10+3)}{4} = \frac{10 \cdot 11 \cdot 12 \cdot 13}{4}$ $\sum_{k=1}^{10} k(k+1)(k+2) = \frac{17160}{4} = 4290$ Now, substitute this back into the sum of reciprocals: $\sum_{k=1}^{10} \frac{1}{a_k} = \frac{1}{4} (4290) = \frac{4290}{4}$

Step 3: Calculate $28 \sum_{k=1}^{10} \frac{1}{a_k}$ and find its prime factorization. We are given that $28 \sum_{k=1}^{10} \frac{1}{a_k} = p_1 p_2 p_3 \ldots p_m$. Substitute the value of the sum from Step 2: $28 \sum_{k=1}^{10} \frac{1}{a_k} = 28 \cdot \frac{4290}{4}$ $28 \sum_{k=1}^{10} \frac{1}{a_k} = 7 \cdot 4290$ Now, we find the prime factorization of $7 \cdot 4290$: $7 \cdot 4290 = 7 \cdot (10 \cdot 429) = 7 \cdot (2 \cdot 5) \cdot (3 \cdot 143) = 7 \cdot 2 \cdot 5 \cdot 3 \cdot (11 \cdot 13)$ Rearranging the prime factors in ascending order: $28 \sum_{k=1}^{10} \frac{1}{a_k} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ The prime numbers $p_1, p_2, \ldots, p_m$ are the distinct prime factors. In this case, the prime factors are 2, 3, 5, 7, 11, and 13. Thus, $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, $p_4 = 7$, $p_5 = 11$, and $p_6 = 13$. The number of these prime factors is $m=6$.

Common Mistakes & Tips

• Incorrectly calculating $S_{n-1}$: Ensure that when substituting $n-1$ into the expression for $S_n$, all occurrences of $n$ are replaced by $n-1$.
• Forgetting to check $a_1$: While the formula $a_n = S_n - S_{n-1}$ is generally for $n \ge 2$, it's crucial to verify if the derived formula for $a_n$ also holds for $n=1$ by comparing it with $a_1 = S_1$.
• Errors in prime factorization: Double-check the prime factorization of the final number to ensure all factors are indeed prime and that no factors are missed.

Summary

The problem required us to first find the general term $a_n$ of the sequence using the given sum $S_n$. This was achieved by calculating $a_n = S_n - S_{n-1}$, leading to $a_n = \frac{4}{n(n+1)(n+2)}$. Next, we computed the sum of the reciprocals of the first 10 terms, $\sum_{k=1}^{10} \frac{1}{a_k}$, by utilizing the formula for the sum of products of consecutive integers, which resulted in $\frac{4290}{4}$. Finally, we multiplied this sum by 28 and performed a prime factorization of the result, $28 \cdot \frac{4290}{4} = 7 \cdot 4290 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$. The number of distinct prime factors, $m$, was found to be 6.

The final answer is \boxed{6} which corresponds to option (C).