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Math Help - Mathematical Induction Problem 2 (IB Math HL)

  1. #1
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    Question Mathematical Induction Problem 2 (IB Math HL)

    Hello.
    I have another question about mathematical induction. I have problem dealing with factorial notation in mathematical induction problems. Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them.
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  2. #2
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    Re: Mathematical Induction Problem 2 (IB Math HL)

    Quote Originally Posted by alireza1992 View Post
    Hello.
    I have another question about mathematical induction. I have problem dealing with factorial notation in mathematical induction problems. Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them.
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    You wish to prove \displaystyle \begin{align*} 1 \cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + n \cdot n! = (n + 1)! - 1 \end{align*}.

    Base step, where \displaystyle \begin{align*} n = 1 \end{align*}:

    \displaystyle \begin{align*} LHS &= 1 \cdot 1! \\ &= 1 \\ \\ RHS &= (1 + 1)! - 1 \\ &= 1 \\ &= LHS  \end{align*}


    Inductive step, assume the statement \displaystyle \begin{align*} 1 \cdot 1! + 2\cdot 2! + 3 \cdot 3! + \dots + n \cdot n! = (n + 1)! - 1 \end{align*} holds true for \displaystyle \begin{align*} n = k \end{align*}, and use this to show it holds true for \displaystyle \begin{align*} n = k + 1 \end{align*}, i.e. show that \displaystyle \begin{align*} 1 \cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + k\cdot k! + (k + 1)\cdot (k + 1)! = (k + 2)! - 1 \end{align*}.

    When we let \displaystyle \begin{align*} n = k + 1 \end{align*} we have

    \displaystyle \begin{align*} LHS &= 1 \cdot 1! + 2 \cdot 2! + 3\cdot 3! + \dots + k \cdot k! + (k + 1)\cdot (k + 1)! \\ &= (k + 1)! - 1 + (k + 1)\cdot (k + 1)! \\ &= (k + 1)! + (k + 1)(k + 1)! - 1 \\ &= (k + 1)!( 1 + k + 1 ) - 1 \\ &= (k + 1)!( k + 2) - 1 \\ &= (k + 2)! - 1 \\ &= RHS  \end{align*}

    Q.E.D.
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    Re: Mathematical Induction Problem 2 (IB Math HL)

    Quote Originally Posted by alireza1992 View Post
    Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them.
    For the first one,

    Base case: 1\times1! = 1 = (1+1)! - 1.

    Now, assume that the statement is true for n = k. So, we have

    1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! = (k+1)! - 1

    \Rightarrow1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! + (k+1)(k+1)!

    . . . . = (k+1)! - 1 + (k+1)(k+1)!

    \Rightarrow1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! + (k+1)(k+1)!

    . . . . = (k+1)! + (k+1)(k+1)! - 1

    . . . . = (k+1)!\left[1 + (k+1)\right] - 1

    . . . . = (k+1)!(k+2) - 1

    . . . . = (k+2)! - 1
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    Re: Mathematical Induction Problem 2 (IB Math HL)

    Thanks a lot!
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    Re: Mathematical Induction Problem 2 (IB Math HL)

    Proof to part b:

    Base: Where n=1:
    LHS=1/2!
    RHS=((1+1)!-1)/(1+1)!=(2!-1)/2!=1/2!=LHS

    Now we assume the statement holds true for n=k, and show that it holds true for n=k+1
    So we have
    1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!=((k+1)!-1)/(k+1)!
    And want to show that
    1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!+((k+1))/(k+2)!=((k+2)!-1)/(k+2)!

    LHS=1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!+((k+1))/(k+2)!

    =((k+1)!-1)/(k+1)!+((k+1))/(k+2)!

    =((k+2)!(k+1)!-(k+2)!+(k+1)(k+1)!)/(k+1)!(k+2)!

    =((k+2)!(k+1)!-(k+2)(k+1)!+(k+1)(k+1)!)/(k+1)!(k+2)!

    =((k+2)!-(k+2)+(k+1))/(k+2)!

    =((k+2)!-k+k-2+1)/(k+2)!

    =((k+2)!-1)/(k+2)!

    Q.E.D.
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