Finding the Product of Sequences a_n and b_n in Terms of x, y, and b_n

[utkarshakash]

Homework Statement

{a_n} and {b_n} are two sequences given by a_n = (x)^{1/2^n}+(y)^{1/2^n} and
b_n = (x)^{1/2^n}-(y)^{1/2^n}. Then find a_1 a_2 a_3 ... a_n in terms of x,y and b_n

Homework Equations

The Attempt at a Solution

I tried substituting n=1,2,3 and so on for a few terms of a_n but really couldn't see anything important. I tried finding x and y instead in terms of a_n and b_n and got this

x =\left( \dfrac{a_n + b_n}{2} \right) ^{2^n} \\<br /> y = \left( \dfrac{a_n - b_n}{2} \right) ^{2^n}

 

[Saitama]

Homework Statement

{a_n} and {b_n} are two sequences given by a_n = (x)^{1/2^n}+(y)^{1/2^n} and
b_n = (x)^{1/2^n}-(y)^{1/2^n}. Then find a_1 a_2 a_3 ... a_n in terms of x,y and b_n

Homework Equations

The Attempt at a Solution

I tried substituting n=1,2,3 and so on for a few terms of a_n but really couldn't see anything important. I tried finding x and y instead in terms of a_n and b_n and got this

x =\left( \dfrac{a_n + b_n}{2} \right) ^{2^n} \\<br /> y = \left( \dfrac{a_n - b_n}{2} \right) ^{2^n}

Observe that
$$b_n=(x)^{1/2^n}-(y)^{1/2^n}=((x)^{1/2^{n+1}})^2-((y)^{1/2^{n+1}})^2$$
That should give you a hint. :)

 

[utkarshakash]

Observe that
$$b_n=(x)^{1/2^n}-(y)^{1/2^n}=((x)^{1/2^{n+1}})^2-((y)^{1/2^{n+1}})^2$$
That should give you a hint. :)

I can only think of using the identity x^2 - y^2 = (x+y)(x-y). But this does not help here :(

 

[Saitama]

I can only think of using the identity x^2 - y^2 = (x+y)(x-y). But this does not help here :(

It does help. Do you see that you should get $b_n=b_{n+1}a_{n+1}$ from your identity?

 

[utkarshakash]

It does help. Do you see that you should get $b_n=b_{n+1}a_{n+1}$ from your identity?

Nice approach Thanks!