If the sequence x_1, x_2, x_3, …, x_n, ... is such that x_1

Question

If the sequence  x_1 ,  x_2 ,  x_3 , …,  x_n , … is such that  x_1  = 3 and  x_{n+1}  = 2 x_n  –1 for n  \geq  1, then  x_{20}  –  x_{19} =

A.  2^{19} 
B.  2^{20} 
C.  2^{21} 
D.  2^{20}  - 1
E.  2^{21}  - 1

Bunuel · Accepted Answer

If the sequence  x_1 ,  x_2 ,  x_3 , ...,  x_n , ... is such that  x_1 = 3  and  x_{n+1} = 2x_n - 1  for  n\geq1 , then  x_{20} - x_{19}  equals which of the following?

A.  2^{19}

B.  2^{20}

C.  2^{21}

D.  2^{20} - 1

E.  2^{21} - 1

We have the sequence  x_1 ,  x_2 ,  x_3 , …,  x_n, , … 
 x_1=3  and  x_{n+1}=2x_n - 1  for  n\geq1 .

If you notice there is a specific pattern in it:
 x_1=3=2^1+1 
 x_2=2x_1-1=5=2^2+1 
 x_3=2x_2-1=9=2^3+1 
...
 x_n=2^n+1

So,  x_{20}=2^{20}+1  and  x_{19}=2^{19}+1 .

x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}

Answer: A.

Hope it helps.

shrouded1 · Answer

x_{n+1}=2x_n - 1 
 x_{n+2}=2*(2x_n - 1) - 1=2^2x_n - (1 + 2) 
 x_{n+3}=2*(2^2x_n - (1+2)) - 1=2^3x_n - (1 + 2 + 2^2) 
 \vdots 
 x_{n+k}=2^kx_n - (1+2+...2^{k-1})=2^k*x_n - (2^k-1)

Let n=1

x_{k+1} = 2^k*3 - (2^k - 1) = 2^{k+1} + 1

So :

x_{20} - x_{19} = 2^{20} - 1 - 2^{19} + 1 = 2^{19}

TimeTraveller · Answer

x1 = 3
x2 = 2(3)-1 = 5
x3 = 2(5)-1 = 9
x4 = 2(9)-1 = 17

we can notice the general pattern in this series:
(x2-x1) = 2 = 2^1
(x3-x2)= 4 = 2^2
(x4-x3) = 8 = 2^3... and so on.

So, (x^20 - x^19) = 2^(19). Ans (A).

TeamGMATIFY · Answer

x_1  = 3
 x_2  = 2 x_1  –1 = 5
 x_3  = 2 x_2  –1 = 9
 x_4  = 2 x_3  –1 = 17
 x_5  = 2 x_4  –1 = 33

On trying to find a series, this translates to 
 x_1  =  2^1 + 1 
 x_2  =  2^2 + 1 
 x_3  =  2^3 + 1 
 x_4  =  2^4 + 1 
 x_5  =  2^5 + 1

Hence  x_{20}  =  2^{20} + 1  and  x_{19}  = 2^{19} + 1

x_{20}  -  x_{19}  = ( 2^{20}  + 1) - ( 2^{19}  + 1) =  2^{20}  -  2^{19}  = 2^{19}  (2 - 1) =  2^{19} 
Option A

bumpbot · Answer

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If the sequence x_1, x_2, x_3, …, x_n, ... is such that x_1

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