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December 17th, 2018, 05:36 AM  #1 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0 
x(0)=1 x(n+1)=2x(n)+1 S=x(0)+x(1)+...+x(n) S(2018)=? I know that the sequence from x(n+1) is 1,3,7,15,31. Now what do I do? How do I find x(n)? Last edited by skipjack; December 17th, 2018 at 10:10 AM. 
December 17th, 2018, 06:18 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,927 Thanks: 1524  Quote:
$\displaystyle S_n = \sum_{k=0}^n (2^k  1) = \sum_{k=0}^n 2^k  \sum_{k=0}^n 1 = \sum_{k=0}^n 2^k  n$ note $\displaystyle \sum_{k=0}^n 2^k = 2^0 + 2^1 + 2^2 + \, ... = 1+2+4+\, ...$ whose sequence of partial sums is $ \{1,3,7,15, \, ... \} = x_n$ so, $\displaystyle S_n = \sum_{k=0}^n 2^k  n = x_n  n$ $S_{2018} = x_{2018}  2018 = (2^{2018}1) 2018 = 2^{2018}  2019$  
December 17th, 2018, 06:27 AM  #3 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  
December 17th, 2018, 06:46 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,927 Thanks: 1524  
December 17th, 2018, 07:14 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,927 Thanks: 1524 
Ok ... I messed up the exponent by starting the index with 1 instead of 0 $\{1,3,7,15,\, ... \} = 2^{n+1}  1$ $\displaystyle S_n = \sum_{k=0}^n 2^{n+1}  1 = \sum_{k=0}^n 2^{n+1}  \sum_{k=0}^n 1 = 2 \sum_{k=0}^n 2^k  (n+1) = 2(2^{n+1}  1)  (n+1) = 2^{n+2}  2  n  1 = 2^{n+2}  (n+3)$ $S_{2018} = 2^{2018+2}  (2018 + 3) = 2^{2020}  2021$ 
December 18th, 2018, 11:14 AM  #6 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0 
@skeeter How does Sum(2^k, k = 0 .. n) =2^(n+1)1?

December 18th, 2018, 11:42 AM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,927 Thanks: 1524  $\displaystyle \sum_{k=0}^n 2^k = 2^0 + 2^1 + 2^2 + 2^3 + \, ... \, + 2^n$ note the partial sums ... $\displaystyle \sum_{k=0}^0 2^k = 2^0 = 1 = 2^1  1$ $\displaystyle \sum_{k=0}^1 2^k = 2^0+2^1 = 3 = 2^2  1$ $\displaystyle \sum_{k=0}^2 2^k = 2^0+2^1+2^2 = 7 = 2^3  1$ $\displaystyle \sum_{k=0}^3 2^k =2^0+2^1+2^2+2^3 = 15 = 2^4  1$ ... $\displaystyle \sum_{k=0}^n 2^k = 2^{n+1}  1$ 
December 18th, 2018, 12:02 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,654 Thanks: 2087 
$\displaystyle \sum_{k=0}^n 2^k = 2\sum_{k=0}^n 2^k  \sum_{k=0}^n 2^k = \sum_{k=1}^{n+1} 2^k  \sum_{k=0}^n 2^k = 2^{n+1}  1$

December 18th, 2018, 01:14 PM  #9 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff.  

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