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Math Help - Proof by Induction ?

  1. #1
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    Wink Proof by Induction ?

    Hi,

    I cannot believe I am having to ask for help, but I have left this far too late and have a ton of work to do. Can someone solve the following problem for me ? Your assistance will be greatly appreciated. Perhaps some notes on how you solved it. Thanks.

    Prove the following by induction:

    n

    ---
    \
    /
    i(i+1)(i+2)=1/4 n(n+1)(n+2)(n+3)
    ---

    i=1


    Please excuse my poor attempt at creating a summation symbol.
    The 1/4 is supposed to represtent 1 quarter.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Charro View Post
    Hi,

    I cannot believe I am having to ask for help, but I have left this far too late and have a ton of work to do. Can someone solve the following problem for me ? Your assistance will be greatly appreciated. Perhaps some notes on how you solved it. Thanks.

    Prove the following by induction:

    n

    ---
    \
    / i(i+1)(i+2)=1/4 n(n+1)(n+2)(n+3)
    ---
    i=1


    Please excuse my poor attempt at creating a summation symbol.
    The 1/4 is supposed to represtent 1 quarter.
    We will prove, by induction that for all n \ge 1 that the statement S(n):

    <br />
\sum_{i=1}^n i (i+1)(i+2) = \frac{ n(n+1)(n+2)(n+3)}{4}<br />

    is true.

    First when n=1 the left hand side is 1 \times 2 \times 3=6, and the right hand side is
    (1/4) \times 1\times 2\times 3 \times 4 = 6, so it is true for n=1

    Now suppose that for some k \ge 1 that the S(k) holds, then:

    <br />
\sum_{i=1}^{k+1} i (i+1)(i+2) = \left[\sum_{i=1}^{k} i (i+1)(i+2)\right] + (k+1)(k+2)(k+3) <br />

    but by supposition the square bracket is equal to:

     \frac{ k(k+1)(k+2)(k+3)}{4} ,

    so:

    <br />
\sum_{i=1}^{k+1} i (i+1)(i+2) = \left[ \frac{ k(k+1)(k+2)(k+3)}{4} \right] + (k+1)(k+2)(k+3) = <br />
\frac{ (k+1)(k+2)(k+3)(k+4)}{4} <br />

    Which is S(k+1).

    To summarise, we have shown that S(1) is true, and that if S(k) is true then so is S(k+1), so in short we
    have proven by mathematical induction that S(n) holds for all n \ge 1

    RonL
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  3. #3
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    Hello, Charro!

    Prove by induction:

    \sum^n_1 i(i+1)(i+2) \;=\;\frac{1}{4}n(n+1)(n+2)(n+3)

    Verify S(1)
    Does . 1(1+1)(1+2) \,=\,\frac{1}{4}(1)(2)(3)(4) ? . . . yes!


    Assume S(k) is true: . \sum^k_1 i(i+1)(i+2) \:=\:\frac{1}{4}k(k+1)(k+2)(k+3)

    Add (k+1)(k+2)(k+3) to both sides:
    . . \sum^k_1i(i+1)(i+2)\;+\;(k+1)(k+2)(k+3) \:=\:\frac{1}{4}k(k+1)(k+2)(k+3)\;+\;(k+1)(k+2)(k+  3)

    The left side is: . \sum^{k+1}_1i(i+1)(i+2)

    Factor the right side: . (k+1)(k+2)(k+3)\left[\frac{1}{4}k + 1\right] \;=\;(k+1)(k+2)(k+3)\left(\frac{k+4}{4}\right)


    The equation becomes: . \sum^{k+1}_1i(i+1)(i+2) \;=\;\frac{1}{4}(k+1)(k+2)(k+3)(k+4)


    And we have established S(k+1) . . . The inductive proof is complete.

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  4. #4
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    Many thanks guys,

    You've inspired me to persevere with this assignment. For some reason this term seems to be falling appart at the seems (done extremely well up to now). Mismanagement of time by myself is mainly to blame :-(

    I'm going to have to go over this later so that I gain a good understanding of how it works.

    I have heard it said that an understanding of mathematics is not really neccessary in life, but they are so wrong. I believe that the more you understand about mathematics the better prepared you will be to solve many real world problems. Obviously not everything, unless you throw in the understanding of logic.

    Thanks again,

    Charro (just incase you are wondering about the name, it is a film and character played by Elvis Presley, the only serious role I am awear of)
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  5. #5
    Member >_<SHY_GUY>_<'s Avatar
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    Exclamation I don't get Proof by Induction

    Hi, im taking an algebra II course that teaches both Algebra A and B in one semester. one thing i don't understand that we have been going through is proof by induction. i dont get the steps, i dont understand it. here are some equations that we reviewed in which i didn't understand:

    1. 2+4+6+8+...+2n= n(n+1)

    2. 1+5+9+13+...+(4n-3)= n(2n-1)

    3. 3+8+13+18+...+(5n-2)= n/2 (5n+1)

    4. 1+2^2+2^3+...+2^n-1= 2^n-1 [^ is used to express in exponents since i dont know how to put it here]

    5. 1^3+2^3+3^3+4^3+...+n^3= n^2 (n+1)^2/ 4

    i was wondering if you can help me because after the teacher doing these, i could not understand how she got (k+1) in some of the equations and how to use it. i am in dieing need of help and i would greatly appreciate it if you can help me
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  6. #6
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    I have moved Shy Guy's post to a new thread.

    -Dan
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