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Math Help - Mathematical induction: sum of odd numbers.

  1. #1
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    Mathematical induction: sum of odd numbers.

    by Set Theory PMI

    Prove: for all Integers n greater than or = to 1.

    1 + 3 + 5 + ... (2n - 1 ) = n^2

    My initial solution but stuck as i went along the way...

    ------------------------------------------------------------------------------------------
    i. when n = 1, then it is true
    ii. if P (k) is true then P( k +1) is true...

    im stuck then here please help i dont know what to do here nex. thanks a lot


    1 + 3 + 5 +, ...(2n - 1) =n^2
    then it is true
    Last edited by mr fantastic; June 25th 2011 at 02:18 PM. Reason: Re-titled.
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    Re: the principle of mathematical induction... can anybody help me?

    Base step: When n = 1, you have

    LHS = 1

    RHS = 1^2 = 1 = LHS.

    Base step is proven.


    Inductive step: Assume P(k) is true, i.e. that 1 + 3 + 5 + ... + (2k-1) = k^2. Using this you need to show that P(k+1) is true.

    For P(k + 1) you are trying to show 1 + 3 + 5 + ... + (2k-1) + (2k+1) = (k + 1)^2.

    LHS = 1 + 3 + 5 + ... + (2k - 1) + (2k + 1)

    = k^2 + 2k + 1 since 1 + 3 + 5 + ... + (2k - 1) = k^2

    = k^2 + k + k + 1

    = k(k + 1) + 1(k + 1)

    = (k + 1)(k + 1)

    = (k + 1)^2

    = RHS.

    So the Inductive step is proven.
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    Re: the principle of mathematical induction... can anybody help me?

    Quote Originally Posted by rcs View Post
    by Set Theory PMI

    Prove: for all Integers n greater than or = to 1.

    1 + 3 + 5 + ... (2n - 1 ) = n^2

    My initial solution but stuck as i went along the way...

    ------------------------------------------------------------------------------------------
    i. when n = 1, then it is true
    ii. if P (k) is true then P( k +1) is true...

    im stuck then here please help i dont know what to do here nex. thanks a lot


    1 + 3 + 5 +, ...(2n - 1) =n^2
    then it is true
    If you are stuck on the induction step,
    then you might be wondering what you need to prove.

    We want to show that the formula being true for "any" n
    causes the formula to be true for the "next" n.

    Let these successive values of n be termed "k" and "k+1".

    Hence we have the following propositions

    P(k)

    1+3+5+....+(2k-1)=k^2

    P(k+1)

    1+3+5+....+[2(k+1)-1]=(k+1)^2\;\;?

    We aim to show that IF P(k) is true, THEN P(k+1) is also true.

    If P(k) is true, then P(k+1) becomes

    \left(1+3+5+.....+2k-1\right)+2k+1=k^2+2k+1\;\;?

    \Rightarrow\ k^2+2k+1=k^2+2k+1

    Hence P(k+1) IS true, IF P(k) is true.
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  4. #4
    rcs
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    Thank you

    thank you so much God BLess You are truly amazing.
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    rcs
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    RHS and LHS meant?

    what does RHS and LHS meant for sir Prove It?

    thanks
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    Re: RHS and LHS meant?

    Quote Originally Posted by rcs View Post
    what does RHS and LHS meant for sir Prove It?

    thanks
    Left Hand Side and Right Hand Side. This is standard notation.
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    Re: Mathematical induction: sum of odd numbers.

    Quote Originally Posted by rcs View Post
    by Set Theory PMI

    Prove: for all Integers n greater than or = to 1.

    1 + 3 + 5 + ... (2n - 1 ) = n^2

    My initial solution but stuck as i went along the way...

    ------------------------------------------------------------------------------------------
    i. when n = 1, then it is true
    ii. if P (k) is true then P( k +1) is true...

    im stuck then here please help i dont know what to do here nex. thanks a lot


    1 + 3 + 5 +, ...(2n - 1) =n^2
    then it is true
    Without induction:

    1 + 3 + 5 + 7 + 9 +... (2n-9)+(2n-7)+(2n-5)+(2n-3)+(2n - 1 )


    Observe that:

    1+(2n-1)=2n

    1+(2n-1)=2n

    3+(2n-3)=2n

    5+(2n-5)=2n

    7+(2n-7)=2n

    .
    .
    .

    The sum of all above is:

    2n+2n+2n+2n+....

    but how many times 2n appears?
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    Re: Mathematical induction: sum of odd numbers.

    it is infinitely many
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    Re: Mathematical induction: sum of odd numbers.

    \begin{aligned} &  S  = \sum_{1 \le k \le n}\left(2k-1\right) =  \sum_{1 \le  n-k \le n}\left[2(n-k)-1\right] = \sum_{1-n \le -k \le 0}\left(2n-2k-1\right) \\& = \sum_{0 \le k \le n-1}\left(2n-2k-1\right) = (n-1)-(2n-2n-1)+\sum_{1 \le k \le n}\left(2n-2k-1\right). \\& \begin{aligned}\therefore ~ 2S & =  2n+\sum_{1 \le k \le n}(2k-1+2n-2k-1) = 2n+\sum_{1 \le k \le n}(2n-2) \\& = 2n+(2n-2) \sum_{1 \le k \le n}(1) = 2n+2n(n-1) = 2n^2. \\& \therefore ~ S = n^2.\end{aligned} \end{aligned}
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    Re: Mathematical induction: sum of odd numbers.

    Quote Originally Posted by rcs View Post
    it is infinitely many
    No! At the original sum there are a finite number of components so if we arrange them in pairs, we'll still have left with a finite number of components.

    Try it with n=7, the sum of seven odd numbers:

    1 + 3 + 5 + 7 + 9 + 11 + 13

    1+13=14
    3+11=14
    5+9=14
    and 7 left alone.

    Try develop this theory of arithmetic summation by your own...



    Also read here:

    How To Be A Little Gauss
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    Re: Mathematical induction: sum of odd numbers.

    Quote Originally Posted by rcs View Post
    by Set Theory PMI

    Prove: for all Integers n greater than or = to 1.

    1 + 3 + 5 + ... (2n - 1 ) = n^2

    My initial solution but stuck as i went along the way...

    ------------------------------------------------------------------------------------------
    i. when n = 1, then it is true
    ii. if P (k) is true then P( k +1) is true...

    im stuck then here please help i dont know what to do here nex. thanks a lot


    1 + 3 + 5 +, ...(2n - 1) =n^2
    then it is true
    check if it is true for n=1
    1 = 1^2
    1 = 1

    Assume it is true for n=k
    1 + 3 + 5 + ....(2k - 1) = k^2
    Check if it is true for n= k + 1
    1 + 3 + 5 + ....(2k -1) + (2(k+1) - 1) = (k+1)^2
    You can replace 1 + 3 +5 +....(2k-1) with k^2
    k^2 + (2(k+1) - 1) = (k+1)^2
    k^2 + (2k +2 - 1) = (k+1)^2
    k^2 + 2k + 1 = (k+1)^2
    (k+1)(k+1) = (k+1)^2
    (k+1)^2 = (k+1)^2

    Proven!
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