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Math Help - Mathematical Induction (2)

  1. #1
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    Mathematical Induction (2)

    Q: Given that f(n) = (1/2!) + (2/3!) + 3/4!) +...+ [n/(n+1)!], where n is a positve integer, find the values of f(1), f(2) and f(3). hence make a conjecture for a forumla for f(n) in terms of n and prove it using mathematical induction.

    Thank you in advance!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

     f(1)=\boxed{\frac{1}{2}} (note that 2!=2)
     f(2)=f(1)+\frac{2}{3\cdot 2\cdot1}=\frac{1}{2}=\frac{1}{2}+\frac{1}{3}=\boxe  d{\frac{5}{6}} (note that 3!=6\\)
     f(3)=f(2)+\frac{3}{4\cdot3\cdot2\cdot1}=\frac{5}{6  }+\frac{1}{8}=\frac{46}{48}=\boxed{\frac{23}{24}} (note that 4!=24)

    Can you guess what has to be shown ?
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  3. #3
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    Hello, Tangera!

    This is quite a task . . .


    Given that: . f(n) \:= \:\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} +\hdots + \frac{n}{(n+1)!} .where n is a positve integer.

    (a) Find the values of f(1),\:f(2)\text{ and }f(3).

    (b) Make a conjecture for a formula for f(n) in terms of n

    (c) Prove it using mathematical induction.

    (a) \quad f(1) \:=\: \frac{1}{2}\qquad f(2) \:=\:\frac{5}{6}\qquad f(3) \:=\:  \frac{23}{24}



    (b) The denominator seems to be: (n+1)!
    . . .The numerator seems to be "one less": (n+1)!-1

    The function seems to be: . f(n) \;=\;\frac{(n+1)!-1}{(n+1)!}



    (c)\;\;S(n)\!:\;\;\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \hdots + \frac{n}{(n+1)!} \;=\;\frac{(n+1)!-1}{(n+1)!}

    Verify S(1)\!:\;\;\frac{1}{2!} \:=\:\frac{2!-1}{2!} \:=\:\frac{1}{2} . . . True

    Assume S(k)\!:\;\;\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \hdots + \frac{k}{(k+1)!} \:=\:\frac{(k+1)!-1}{(k+1)!}


    Add \frac{k+1}{(k+2)!} to both sides:

    . . \underbrace{\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \hdots + \frac{k+1}{(k+2)!}} \;=\;\frac{(k+1)!-1}{(k+1)!} +\frac{k+1}{(k+2)!}

    The left side is the left side of S(k+1).
    We must show that the right side is the right side of S(k+1)\!:\;\;\frac{(k+2)!-1}{(k+2)!}

    The right side is: . {\color{blue}\frac{k+2}{k+2}}\!\cdot\!\frac{(k+1)!-1}{(k+1)!} + \frac{k+1}{(k+2)!} \;=\;\frac{(k+2)[(k+1)! - 1] + (k+1)}{(k+2)!}

    . . = \;\frac{\overbrace{(k+2)(k+1)!}^{\text{This is }(k+2)!}\, \overbrace{- (k+2) + (k+1)}^{\text{This is -1}}}{(k+2)!} \;=\;\frac{(k+2)!-1}{(k+2)!} . . . There!

    The inductive proof is complete.

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  4. #4
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    Hello Soroban and flyingsquirrel! Thank you very much for your help!
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