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 December 17th, 2018, 05:36 AM #1 Member   Joined: Aug 2016 From: Romania Posts: 30 Thanks: 1 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
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Quote:
 Originally Posted by alex77 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)?
from observation, $x_n = 2^n-1$

$\displaystyle S_n = \sum_{k=0}^n (2^k - 1) = \sum_{k=0}^n 2^k - \sum_{k=0}^n 1 = \sum_{k=0}^n 2^k - n$

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
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 Originally Posted by skeeter $S_{2018} = x_{2018} - 2018 = (2^{2018}-1) -2018 = 2^{2018} - 2019$
@skeeter The answer in my book was 2^2020-2021.

December 17th, 2018, 06:46 AM   #4
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 Originally Posted by alex77 @skeeter The answer in my book was 2^2020-2021.
I'll have to go back over it and see if I can find an error ...

 December 17th, 2018, 07:14 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 Ok ... I messed up the exponent by starting the index with 1 instead of 0 $\{1,3,7,15,\, ... \} = 2^{n+1} - 1$ $\displaystyle S_n = \sum_{k=0}^n 2^{n+1} - 1 = \sum_{k=0}^n 2^{n+1} - \sum_{k=0}^n 1 = 2 \sum_{k=0}^n 2^k - (n+1) = 2(2^{n+1} - 1) - (n+1) = 2^{n+2} - 2 - n - 1 = 2^{n+2} - (n+3)$ $S_{2018} = 2^{2018+2} - (2018 + 3) = 2^{2020} - 2021$ Thanks from topsquark
 December 18th, 2018, 11:14 AM #6 Member   Joined: Aug 2016 From: Romania Posts: 30 Thanks: 1 @skeeter How does Sum(2^k, k = 0 .. n) =2^(n+1)-1?
December 18th, 2018, 11:42 AM   #7
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 Originally Posted by alex77 @skeeter How does Sum(2^k, k = 0 .. n) =2^(n+1)-1?
$\displaystyle \sum_{k=0}^n 2^k = 2^0 + 2^1 + 2^2 + 2^3 + \, ... \, + 2^n$

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: 20,942 Thanks: 2210 $\displaystyle \sum_{k=0}^n 2^k = 2\sum_{k=0}^n 2^k - \sum_{k=0}^n 2^k = \sum_{k=1}^{n+1} 2^k - \sum_{k=0}^n 2^k = 2^{n+1} - 1$ Thanks from topsquark
December 18th, 2018, 01:14 PM   #9
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Quote:
 Originally Posted by skipjack $\displaystyle \sum_{k=0}^n 2^k = 2\sum_{k=0}^n 2^k - \sum_{k=0}^n 2^k = \sum_{k=1}^{n+1} 2^k - \sum_{k=0}^n 2^k = 2^{n+1} - 1$
Nice. I was going to do an induction on it, but yours is much quicker.

-Dan

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