Find the product of this term

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


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

[itex]x =\left( \dfrac{a_n + b_n}{2} \right) ^{2^n} \\
y = \left( \dfrac{a_n - b_n}{2} \right) ^{2^n}[/itex]
 

Answers and Replies

  • #2
3,812
92

Homework Statement


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

[itex]x =\left( \dfrac{a_n + b_n}{2} \right) ^{2^n} \\
y = \left( \dfrac{a_n - b_n}{2} \right) ^{2^n}[/itex]

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. :)
 
  • #3
utkarshakash
Gold Member
855
13
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 :(
 
  • #4
3,812
92
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?
 
  • #5
utkarshakash
Gold Member
855
13
It does help. Do you see that you should get ##b_n=b_{n+1}a_{n+1}## from your identity?

Nice approach Thanks!
 

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