Series

Algebra Level 4

If S n = ∑ m = 1 n 2 m − 1 \displaystyle S_{n}=\sum_{m=1}^{n}2^{m-1} , compute ∑ m = 1 1729 log ⁡ 2 m ( S m + 1 ) \displaystyle \sum_{m=1}^{1729}\log_{2^{m}}(S_{m}+1) .


The answer is 1729.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Prakhar Bindal
Jul 16, 2015

We Can Simply Evaluate The Sum Sn Because Its A Increasing Geometric Progession Upto "n" terms .

Apply The Formula Of GP We Get the sum to be 2^m -1 . Substitute this value into the logarithmic series

We Get That The Number And Base Of Logarithm both are same . hence the value of log will be 1 for any real m .

Hence the sum will boil down to 1+1+1........1729 times = 1729

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