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Math Help - Homework problem

  1. #1
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    Homework problem

    We know that

    \sum_{k=1}^nk=\frac{n(n+1)}{2}

    where n is a natural number. obtain this formula using the identity.

    (k+1)^2-k^2=2k+1

    to obtain the expression

    \sum_{k=1}^n(2k+1) then use a similar approach to obtain the formula

    \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}

    DO NOT USE INDUCTION

    There is the question, I don't even know where to begin :/
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jpatrie View Post
    We know that

    \sum_{k=1}^nk=\frac{n(n+1)}{2}

    where n is a natural number. obtain this formula using the identity.

    (k+1)^2-k^2=2k+1

    to obtain the expression

    \sum_{k=1}^n(2k+1) then use a similar approach to obtain the formula
    This is about telescoping sums:

     \sum_{k=1}^n(2k+1) = \sum_{k=1}^n (k+1)^2 - k^2

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

    Now all but the -1^2 and (n+1)^2 terms cancell, so:

     \sum_{k=1}^n(2k+1) =(n+1)^2-1

    RonL
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  3. #3
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    Hello, jpatrie!

    We know that: . \sum_{k=1}^nk=\frac{n(n+1)}{2}, where n is a natural number.

    Obtain this formula using the identity: . (k+1)^2-k^2\:=\:2k+1

    We have: . (k+1)^2 - k^2 \:=\:2k+1

    Let k = 1,2,3,\hdots, n and "stack" the equations.

    . . \begin{array}{cccc} k=1\!: & 2^2 - 1^2 &=& 2(1) + 1 \\ k=2\!: & 3^2 - 2^2 &=& 2(2) + 1 \\ k=3\!: & 4^2 - 3^2 &=& 2(3) + 1 \\ \vdots & \vdots & & \vdots \\ k=n\!: & (n+1)^2 - n^2 &=& 2n + 1 \end{array}


    Now add the equations ... note that most of the left side cancels out.

    . . (n+1)^2 - 1 \;=\;2\underbrace{(1 + 2 + 3 + \hdots + n)}_{\text{This is }\sum k} + \underbrace{(1 + 1 + 1 + \hdots + 1)}_{n\text{ terms}}

    So we have: . n^2 +2n \;=\;2\sum k + n \quad\Rightarrow\quad 2\sum k \;=\;n^2 + n


    Therefore: . \boxed{\sum k \;=\;\frac{n(n+1)}{2}}




    Then use a similar approach to obtain the formula: . \sum_{k=1}^nk^2\;=\;\frac{n(n+1)(2n+1)}{6}

    We have: . (k+1)^3 - k^3 \;=\;3k^2 + 3k + 1

    Let k = 1,2,3,\cdots, n

    . . \begin{array}{ccccc}k=1\!: & 2^3 - 1^3 &=& 3(1^2) + 3(1) + 1 \\<br />
k=2\!: & 3^3 - 2^3 &=& 3(2^2) + 3(2) + 1 \\<br />
k=3\!: & 4^3 - 3^3 &=& 3(3^2) + 3(3) + 1 \\<br />
\vdots & \vdots & & \vdots \\<br />
k=n\!: & (n+1)^3 - n^3 &=& 3(n^2) + 3(n) + 1 \end{array}


    Add the equations:

    . . (n+1)^3 - 1 \;=\;3\underbrace{\left[1^2 + 2^2 + 3^2 +\hdots+n^2\right]}_{\text{This is }\sum k^2} \:+\: 3\underbrace{\left[1 + 2 + 3 +\hdots+n\right]}_{\text{This is }\sum k} + .  \underbrace{[1 + 1 + 1 + \hdots + 1]}_{n\text{ terms}}

    So we have: . n^3 + 3n^2 + 3n \;=\;3\sum k^2 + 3\cdot\frac{n(n+1)}{2} + n

    . . which simplifies to: . 3\sum k^2 \;=\;\frac{2n^3 + 3n^2 + n}{2} \;=\;\frac{n(n+1)(2n+1)}{2}


    Therefore: . \boxed{\sum k^2 \;=\;\frac{n(n+1)(2n+1)}{6}}

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  4. #4
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    Thanks a lot guys, but can anyone prove it using "recursion", unless those are the methods of recursion.
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