Let $< a_{\mathrm{n}} >$ be a sequence such that $a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$. If $28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$, where $\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$ are the first $\mathrm{m}$ prime numbers, then $\mathrm{m}$ is equal to

Given the sum of the first n terms, $S_n = \frac{n^2+3n}{(n+1)(n+2)}$, we can find the n^th term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms : <br/><br/>So, $a_n = S_n - S_{n-1}$ <br/><br/>Solving, we get : <br/><br/>$a_n = \frac{n^2+3n}{(n+1)(n+2)} - \frac{(n-1)^2+3(n-1)}{n(n+1)}$ <br/><br/>Simplifying further, we find : <br/><br/>$a_n = \frac{4}{n(n+1)(n+2)}$ <br/><br/>Then, we find the reciprocal of $a_n$: <br/><br/>$\frac{1}{a_n} = \frac{n(n+1)(n+2)}{4}$ <br/><br/>Now, we sum this over the first 10 terms : <br/><br/>$$\sum\limits_{k=1}^{10} \frac{1}{a_k} = \sum\limits_{k=1}^{10} \frac{k(k+1)(k+2)}{4}$$ <br/><br/>Evaluating the sum : <br/><br/>$$ \sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} \left[\sum\limits_{k=1}^{10} k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2)\right] $$ <br/><br/>This can be rewritten as the sum of differences : <br/><br/>$$ \sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} [(1 \cdot 2 \cdot 3 \cdot 4 - 0) + (2 \cdot 3 \cdot 4 \cdot 5 - 1 \cdot 2 \cdot 3 \cdot 4) + \cdots + (10 \cdot 11 \cdot 12 \cdot 13 - 9 \cdot 10 \cdot 11 \cdot 12)] $$ <br/><br/>$$ \sum\limits_{k=1}^{10} \frac{1}{a_k} = \frac{1}{16} (10 \cdot 11 \cdot 12 \cdot 13 - 0) = \frac{1}{16} \cdot 17160 $$ <br/><br/>Now, given the condition that : <br/><br/>$28 \sum\limits_{k=1}^{10} \frac{1}{a_k} = p_1p_2p_3...p_m$ <br/><br/>Substituting the sum we've calculated: <br/><br/>$28 \cdot \frac{1}{16} \cdot 17160 = p_1p_2p_3...p_m$ <br/><br/>This simplifies to : <br/><br/>$30030 = p_1p_2p_3...p_m$ <br/><br/>The prime factorization of 30030 is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$, which consists of 6 primes. <br/><br/>Therefore, m is equal to 6.