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Sequence problem

by pbanavara » Sun Nov 23, 2008 5:35 pm
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n 1, then x20 – x19 = ?

I tried using x20 = 2 x19 -1
so x20-x19 = 2x19 - 1 - x19, which is like 2a - 1 -a = a -1 =>x19 - 1

How do you deduce x19 ? It's quite time consuming to go all the way to x1 from x19 .. any clues will be greatly helpful.

Thanks,
Pradeep
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Re: Sequence problem

by logitech » Sun Nov 23, 2008 5:59 pm
pbanavara wrote:If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n 1, then x20 – x19 = ?

I tried using x20 = 2 x19 -1
so x20-x19 = 2x19 - 1 - x19, which is like 2a - 1 -a = a -1 =>x19 - 1

How do you deduce x19 ? It's quite time consuming to go all the way to x1 from x19 .. any clues will be greatly helpful.

Thanks,
Pradeep
X2 = 2X1-1
X3 = 4X1-3
x4 = 8x1-7

So you can see that if you substract 1 from each sides:

x2-1 = 2(x1-1) 2=2^1
x3-1 = 4(x1-1) 4=2^2
x4-1= 8(x1-1) 8=2^3
....

So Xn-1 = 2^(n-1) (x1-1) x-1= (3-1) = 2

.... xn-1 = 2^(n-1) 2

....Xn-1 = 2^n

Xn = 2^n + 1

x20 – x19 = 2^20 +1 - (2^19 +1 )

=2^19
Last edited by logitech on Mon Nov 24, 2008 12:17 am, edited 2 times in total.
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Re: Sequence problem

by pbanavara » Sun Nov 23, 2008 7:04 pm
logitech wrote:
pbanavara wrote:If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n 1, then x20 – x19 = ?

I tried using x20 = 2 x19 -1
so x20-x19 = 2x19 - 1 - x19, which is like 2a - 1 -a = a -1 =>x19 - 1

How do you deduce x19 ? It's quite time consuming to go all the way to x1 from x19 .. any clues will be greatly helpful.

Thanks,
Pradeep
X2 = 2X1-1
X3 = 4X1-3
x4 = 8x1-7

So you can see that if you substract 1 from each sides:

x2-1 = 2(x1-1) 2=2^1
x3-1 = 4(x1-1) 4=2^2
x4-1= 8(x1-1) 8=2^3
....

So Xn-1 = 2^(n-1) (x1-1) x-1= (3-1) = 2

.... xn-1 = 2^(n-1) 2

....Xn-1 = 2^n

x20 – x19 = 2^21 - 2^20

= 2^20(2-1)
= 2^20
The first 3 lines were the key. Thanks much for the solution.

- pradeep
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by 4meonly » Mon Nov 24, 2008 12:02 am
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n 1, then x20 – x19 = ?
if X1 = 3
and we have geometric sequence Xn+1 = 2*Xn – 1
then Xn = (X1) * (2)^n-1 or Xn = (3) * (2)^n-1

So,
X20 = 3 * 2^19
X19 = 3 * 2^18
X20-X19 = 3 * 2^19 - 3 * 2^18 = 3 * 2^18
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by logitech » Mon Nov 24, 2008 12:13 am
4meonly wrote:
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n 1, then x20 – x19 = ?
if X1 = 3
and we have geometric sequence Xn+1 = 2*Xn – 1
then Xn = (X1) * (2)^n-1 or Xn = (3) * (2)^n-1

So,
X20 = 3 * 2^19
X19 = 3 * 2^18
X20-X19 = 3 * 2^19 - 3 * 2^18 = 3 * 2^18
According to your equation:

Xn = (X1) * (2)^n-1

for N=2

X2=3*2 = 6

According to the original equation

x1 = 3 and xn+1 = 2xn – 1

X2=2*3 - 1 = 5

According to my equation:

Xn = 2^n + 1

X2 = 2^2 + 1 = 5

What is the OA for this question ?
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by 4meonly » Mon Nov 24, 2008 1:19 am
Sorry,
xn+1 = 2xn – 1
is not the same Xn = (X1) * (2)^n-1

I overlooked..

But if xn+1 = 2xn – 1
we need not only x1 but also x2

Do you agree?
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by logitech » Mon Nov 24, 2008 1:37 am
4meonly wrote:Sorry,
xn+1 = 2xn – 1
is not the same Xn = (X1) * (2)^n-1

I overlooked..

But if xn+1 = 2xn – 1
we need not only x1 but also x2

Do you agree?
We can find out X2 when we use n=1

x2 = 2x1-1

Did I understand you correctly ?
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by 4meonly » Mon Nov 24, 2008 2:44 am
X2 = 2 * X1-1 = 2 * X0
but we dont know X0

X3 = 2 * X1 = 6
X4 = 2 * X2 but we dont know X2

That is the problem
n is the number is the sequence
we should not meltiply 2 by n
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by jimmiejaz » Mon Nov 24, 2008 5:59 am
Guys how about this approach,

x1=3, x2=5, x3=9......
x2=x1+2
x3=x2+2^2
so....
x20=x19+2^19
Now we have to find out x20-x19
substituting,
=x19+2^19-x19
=2^19

Please correct me if am wrong....
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by logitech » Mon Nov 24, 2008 8:22 am
4meonly wrote:X2 = 2 * X1-1 = 2 * X0
but we dont know X0
I believe that you misread the question. ( or I did )

This is the original equation:

xn+1 = (2Xn) - 1
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by pbanavara » Mon Nov 24, 2008 8:32 am
The OA is 2^19.
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by pbanavara » Mon Nov 24, 2008 8:33 am
logitech wrote:
4meonly wrote:X2 = 2 * X1-1 = 2 * X0
but we dont know X0
I believe that you misread the question. ( or I did )

This is the original equation:

xn+1 = (2Xn) - 1
Logitech - you are right the question is xn+1 = (2xn) - 1
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by pbanavara » Mon Nov 24, 2008 8:35 am
jimmiejaz wrote:Guys how about this approach,

x1=3, x2=5, x3=9......
x2=x1+2
x3=x2+2^2
so....
x20=x19+2^19
Now we have to find out x20-x19
substituting,
=x19+2^19-x19
=2^19

Please correct me if am wrong....
This works as well - Thanks
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by 4meonly » Mon Nov 24, 2008 9:37 am
pbanavara wrote:
logitech wrote:
4meonly wrote:X2 = 2 * X1-1 = 2 * X0
but we dont know X0
I believe that you misread the question. ( or I did )

This is the original equation:

xn+1 = (2Xn) - 1
Logitech - you are right the question is xn+1 = (2xn) - 1
I thought
Xn+1 = 2 * Xn-1
:x

Guys, please do not forget to use ) and (!!!!!
or make a pricture of the problem otherwise you can meslead others!
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Step 1

by JGS » Mon Nov 24, 2008 9:57 am
How do you determine

x2=5 and x3=9?
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