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Thread: 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

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

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

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

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

    (c) Prove it using mathematical induction.

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



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

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



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

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

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


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

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

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

    The right side is: . $\displaystyle {\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)!} $

    . . $\displaystyle = \;\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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