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

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

If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n 1, then x20 – x19 = ?

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

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?

4meonly wrote:X2 = 2 * X1-1 = 2 * X0
but we dont know X0

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

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....

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