Geometric Series Equivalence

May 2013
19
0
Canada
Hello,
Why is this:
1-Divide_Conquer_Feb28March2-2017__pdf__page_9_of_19_.png

Equivalent to this:
1-Divide_Conquer_Feb28March2-2017__pdf__page_9_of_19_.png

Thank you.
 

ChipB

MHF Helper
Jun 2014
305
124
NJ
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

\(\displaystyle \sum _{i=0} ^{k-1} 2^i = 2^k-1\)
 
Jun 2013
1,096
573
Lebanon
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