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∑k=1∞6k(3k−2k)(3k+1−2k+1)= ?\sum _{k=1}^\infty \frac {6^k}{(3^k - 2^k)(3^{k+1} - 2^{k+1})} = \ ?k=1∑∞(3k−2k)(3k+1−2k+1)6k= ?
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