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+$$ n_{d}+N_{d}\cdot (n_{d-1}+N_{d-1}\cdot (n_{d-2}+N_{d-2}\cdot (\cdots +N_{2}n_{1})\cdots )))=\sum _{k=1}^{d}\left(\prod _{\ell =k+1}^{d}N_{\ell }\right)n_{k} $$
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+$$ n_{1}+N_{1}\cdot (n_{2}+N_{2}\cdot (n_{3}+N_{3}\cdot (\cdots +N_{d-1}n_{d})\cdots )))=\sum _{k=1}^{d}\left(\prod _{\ell =1}^{k-1}N_{\ell }\right)n_{k} $$
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