Let $<{\mathrm{a}}_{\mathrm{n}}>$ be a sequence such that ${\mathrm{a}}_1+{\mathrm{a}}_2+......+{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2+3\mathrm{n}}{(\mathrm{n}+1)(\mathrm{n}+2)}$. If $28\sum _{\mathrm{k}=1}^{10}\frac{1}{{\mathrm{a}}_{\mathrm{k}}}={\mathrm{p}}_{{ }_1}{\mathrm{p}}_{{ }_2}{\mathrm{p}}_{{ }_3}....{\mathrm{p}}_{{ }_{\mathrm{m}}},$where ${\mathrm{p}}_{{ }_1},{\mathrm{p}}_{{ }_2},....,{\mathrm{p}}_{{ }_{\mathrm{m}}}$are the first m prime numbers, then m is equal to

Let ${\mathrm{S}}_{\mathrm{n}}={\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+.......+{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2+3\mathrm{n}}{(\mathrm{n}+1)(\mathrm{n}+2)}$
$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}={\mathrm{S}}_{\mathrm{n}}-{\mathrm{S}}_{\mathrm{n}-1}=\frac{{\mathrm{n}}^2+3\mathrm{n}}{(\mathrm{n}+1)(\mathrm{n}+2)}-\frac{{(\mathrm{n}-1)}^2+3(\mathrm{n}-1)}{\mathrm{n}(\mathrm{n}+1)} \\ \Rightarrow {\mathrm{a}}_{\mathrm{n}}=\frac{4}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)} \\ 28\sum _{\mathrm{k}=1}^{10}\frac{1}{{\mathrm{a}}_{\mathrm{k}}}=28\sum _{\mathrm{k}=1}^{10}\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{4} \\ =7\sum _{\mathrm{k}=1}^{10}\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2) \\ =7\sum _{\mathrm{k}=1}^{10}\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)-(\mathrm{k}-1)\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{4} \\ =\frac{7}{4}\left[\left(10\right)\left(11\right)\left(12\right)\left(13\right)\right]=2\times 3\times 5\times 7\times 11\times 13 \\ \therefore \mathrm{m}=6 \end{gathered}$$