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$If \sum_{r = 1}^{n} T_{r} = (3^{n} - 1), then find the sum of \sum_{m = 1}^{n} \frac{1}{T_{m}} .$

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Correct option is B

∑r=1nTr=3n−1⇒Tr=3r-1-3r−1−1=3r−1(3−1)=23r−1⇒1Tr=12×13r−1⇒∑r=1n 1Tr=12∑r=1n 13r−1=121−13n1−13=341−13n

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If ∑r=1n Tr=3n−1, then find the sum of ∑m=1n 1Tm.

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$If \sum_{r = 1}^{n} T_{r} = (3^{n} - 1), then find the sum of \sum_{m = 1}^{n} \frac{1}{T_{m}} .$