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If Sn=∑m=1n2m−1 \displaystyle S_{n}=\sum_{m=1}^{n}2^{m-1}Sn=m=1∑n2m−1, compute ∑m=11729log2m(Sm+1) \displaystyle \sum_{m=1}^{1729}\log_{2^{m}}(S_{m}+1) m=1∑1729log2m(Sm+1).
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