C ChipB MHF Helper Jun 2014 305 124 NJ Mar 20, 2017 #2 Since summations only work with integers, you can replace the log(n) parts with an integer, such as k. The write a proof by induction to show that \(\displaystyle \sum _{i=0} ^{k-1} 2^i = 2^k-1\)

Since summations only work with integers, you can replace the log(n) parts with an integer, such as k. The write a proof by induction to show that \(\displaystyle \sum _{i=0} ^{k-1} 2^i = 2^k-1\)

I Idea Jun 2013 1,096 573 Lebanon Mar 20, 2017 #3 we expect \(\displaystyle \sum _{i=0}^{\text{Log} n-1} 2^i\) to be an integer but \(\displaystyle 2^{\text{Log} n}-1\) is not necessarily an integer

we expect \(\displaystyle \sum _{i=0}^{\text{Log} n-1} 2^i\) to be an integer but \(\displaystyle 2^{\text{Log} n}-1\) is not necessarily an integer