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Math Help - [SOLVED] proof by induction

  1. #1
    princess
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    Exclamation [SOLVED] proof by induction

    proof by induction; you follow the three steps ive got that much but i dont know how to do it when the numbers are squared.
    for example...
    1squared + 2squared + 3 squared + ........... + nsquared = 1/6 (n+1)(2n+1)

    i can complete the first step (prove its true for 1)
    1/6 (1+1)(2+1) = 1/6 x 2 x 3
    = 1
    1squared = 1

    step 2 is to assume that it is true for n=k i think i am getting lost on this step.
    i have reached
    1/6k (k+1)(2k+1) = m (i think that part is right)

    however step three is to prove it is true for n=k+1
    this is completely loosing me and it is due tomorrow im totally lost
    please help me its sooo urgent.

    p.s i dont want the answer just some help if that is possible, im more interested in how the answer is recieved then the actual answer that it gives.
    thanks in advance
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  2. #2
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    The equation editor (LaTeX) is curretly disabled because the site is under repair. Please try to understand this mess.
    ---
    The first part you got correct, you checked n=1.

    The second part is,
    <br />
1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}

    Now, to get the next case you add  (k+1)^2 to both sides, thus,

    1^2+2^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+  1)^2

    Now, add the fraction on the right hand side,
    1^2+2^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)+6(k+1)^  2}{6}
    Thus, the right hand side, can be factored
    \frac{(k+1)[k(2k+1)+6(k+1)]{6}
    Thus,
    \frac{(k+1)[2k^2+k+6k+6]{6}
    Thus,
    \frac{(k+1)(k+2)(2k+3)}{6}
    It works!!!
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by princess View Post
    proof by induction; you follow the three steps ive got that much but i dont know how to do it when the numbers are squared.
    for example...
    1squared + 2squared + 3 squared + ........... + nsquared = 1/6 (n+1)(2n+1)

    i can complete the first step (prove its true for 1)
    1/6 (1+1)(2+1) = 1/6 x 2 x 3
    = 1
    1squared = 1

    step 2 is to assume that it is true for n=k i think i am getting lost on this step.
    i have reached
    1/6k (k+1)(2k+1) = m (i think that part is right)
    If you assume the result true for n=k, you have:

    1^2 + 2^2 + ... + k^2 = (k+1)(2k+1)/6,

    from this you need to show that:

    1^2 + 2^2 + ... + k^2 + (k+1)^2 = (k+2)(2k+5)/6,

    which is what the result would give if n=k+1. You would usually proceed
    by noting by the assumption that:

    1^2 + 2^2 + ... + k^2 + (k+1)^2 = (k+1)(2k+1)/6 + (k+1)^2.

    Then you rearrange the right hand side of this last equation until
    it looks like the right hand side of the previous equation, and then
    this stage is done.

    RonL
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