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(B) Let $S_n = \frac{n^2+3n}{(n+1)(n+2)}$.For $n=1$,$a_1 = S_1 = \frac{1+3}{2 	imes 3} = \frac{4}{6} = \frac{2}{3}$.For $n > 1$,$a_n = S_n - S_{n-1}

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$6$

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(B) Let $S_n = \frac{n^2+3n}{(n+1)(n+2)}$.For $n=1$,$a_1 = S_1 = \frac{1+3}{2 	imes 3} = \frac{4}{6} = \frac{2}{3}$.For $n > 1$,$a_n = S_n - S_{n-1} = \frac{n^2+3n}{(n+1)(n+2)} - \frac{(n-1)^2+3(n-1)}{n(n+1)} = \frac{n^2+3n}{(n+1)(n+2)} - \frac{n^2+n-2}{n(n+1)}$.Simplifying this,we get $a_n = \frac{n(n^2+3n) - (n+2)(n^2+n-2)}{n(n+1)(n+2)} = \frac{n^3+3n^2 - (n^3+2n^2-4)}{n(n+1)(n+2)} = \frac{n^2+4}{n(n+1)(n+2)}$.Wait,re-evaluating the sum: $S_n = \frac{n(n+3)}{(n+1)(n+2)} = 1 - \frac{2}{(n+1)(n+2)}$.Then $a_n = S_n - S_{n-1} = (1 - \frac{2}{(n+1)(n+2)}) - (1 - \frac{2}{n(n+1)}) = \frac{2}{n(n+1)} - \frac{2}{(n+1)(n+2)} = \frac{2(n+2-n)}{n(n+1)(n+2)} = \frac{4}{n(n+1)(n+2)}$.Thus,$\frac{1}{a_k} = \frac{k(k+1)(k+2)}{4}$.Then $28 \sum_{k=1}^{10} \frac{1}{a_k} = 28 \sum_{k=1}^{10} \frac{k(k+1)(k+2)}{4} = 7 \sum_{k=1}^{10} k(k+1)(k+2)$.Using the identity $\sum_{k=1}^n k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$,we get:$7 	imes \frac{10 	imes 11 	imes 12 	imes 13}{4} = 7 	imes 10 	imes 11 	imes 3 	imes 13 = 7 	imes (2 	imes 5) 	imes 11 	imes 3 	imes 13 = 2 	imes 3 	imes 5 	imes 7 	imes 11 	imes 13$.This is the product of the first $6$ prime numbers $(2, 3, 5, 7, 11, 13)$.Therefore,$m = 6$.

Let $\langle a_n \rangle$ be a sequence such that $a_1+a_2+\ldots+a_n = \frac{n^2+3n}{(n+1)(n+2)}$. If $28 \sum_{k=1}^{10} \frac{1}{a_k} = p_1 p_2 p_3 \ldots p_m$,where $p_1, p_2, \ldots, p_m$ are the first $m$ prime numbers,then $m$ is equal to

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