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Show it without using induction?

  1. Oct 18, 2004 #1
    how to show this:

    if f(x) = 2x/(2x-1)

    then show f(n) + f(n)f(n-1) + f(n)f(n-1)f(n-2) + ... + f(n)f(n-1)f(n-2)...f(1) = 2n .


    I have tried to use induction , and I can show it.
    but , I can't show it without using induction.


    can anyone tell how to show it without using induction?

    thank you ! :smile:
     
  2. jcsd
  3. Oct 18, 2004 #2

    matt grime

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    let S(n) be the sum you want.

    Can you relate S(n) to S(n-1) algebraically? Hint, does the LHS of your equation have anything common to every term?
     
  4. Oct 18, 2004 #3

    HallsofIvy

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    Do you see what happens when you write out a few terms of the products?

    [tex]f(1)f(2)= \frac{2}{1}\frac{4}{3}= \frac{(2)(4)}{(1){3}}[/tex]
    [tex]f(1)f(2)f(3)= \frac{2}{1}\frac{4}{3}\frac{6}{5}= \frac{(2)(4)(6)}{(1)(3)(5)[/tex]
    and so on. You get a fraction with something "like" a factorial in both numerator and denominator except that the numerator has only even numbers and the denominator only odd numbers.

    (2)(4)(6)...(2n)= (2)(1)(2)(2)(2)(3)...(2)(n)= 2n n!
    and (1)(3)(5)...(2n+1)= ((1)(2)(3)(4)(5)...(2n)(2n+1))/(2)(4)...(2n))
    = (2n+1)!/(2nn!)
    Does that give you any ideas?
     
  5. Oct 18, 2004 #4

    I find S(n) = f(n)*S(n-1)

    and f(n) = (2^2n)(n!)/(2n+1)!

    then , S(n) = product of (2^2r)(r!)/(2r+1)! from r = 1 to n

    am I right ?

    however i still don't know how to do?
     
  6. Oct 18, 2004 #5

    matt grime

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    No, you're not right.
     
  7. Oct 19, 2004 #6
    Oh! :tongue:

    it should be S(n) = f(n)[1 + S(n-1)]

    but still don't know :confused:

    if it is in this from:

    X(n) = aX(n-1) + b

    i can find X(n) in term of a , b and X(1) ,


    but now the a and the b is changing.................................. :confused:
     
  8. Oct 19, 2004 #7

    matt grime

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    You've found a first order difference equation. As with differential equations you guess a solution by inspection based upon f(n).
     
  9. Oct 20, 2004 #8
    S(n) = f(n)[1 + S(n-1)]

    S(n) - S(n-1) = f(n)[1 + S(n-1)] - S(n-1)
    = f(n) + [f(n) - 1]S(n-1)

    ____________________________________________________

    S(1) = 2/1 = 2

    S(2) = 4/3 + 8/3 = 12/3 = 4

    S(3) = 6/5 + (6*4)/(5*3) + (6*4*2)/(5*3*1)
    = 2*3/5 + 2*4/5 + 2*4*2/5
    = 2[3/5 + 3*(4/5)]
    = 2*3*(5/5) = 6

    S(4)
    ......
    ....
    ...
    ..
    .

    S(2) - S(1) = 2....................
    ........................


    I realy cannot find the solution :cry:
     
    Last edited: Oct 20, 2004
  10. Oct 20, 2004 #9

    matt grime

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    You're given 2n, and asked to *show* it is the solution, so it is perfectly acceptable to check it satisfies the recurrence relation by simple substitution.

    And:

    [tex]S(n) = \frac{2n(1+S(n-1))}{2n-1}[/tex]

    it is clear what kind of answer you should guess from looking at that ie that 2n is what you want since 2n-1 = 2(n-1)+1
     
    Last edited: Oct 20, 2004
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