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Math Help - Sum of cubes.

  1. #1
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    Sum of cubes.

     <br /> <br />
\begin{array}{l}<br />
 prove\;that\; \\ <br />
  \\ <br />
 1^3  + 2^3  + 3^3  + ............ + 100^3  = (\frac{{100 \times 101}}{2})^2  \\ <br />
  \\ <br />
 \end{array}<br /> <br /> <br />
    Last edited by mr fantastic; July 27th 2010 at 01:21 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by fxs12 View Post
     <br /> <br />
\begin{array}{l}<br />
prove\;that\; \\ <br />
\\ <br />
1^3 + 2^3 + 3^3 + ............ + 100^3 = (\frac{{100 \times 101}}{2})^2 \\ <br />
\\ <br />
\end{array}<br /> <br /> <br />
    Use Google to fnd the sum of cubes of formula. Now apply that formula.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    1^3+2^3+3^3+...+n^3=(1^2+2^2+3^2+...+n^2)^2={(n(n+ 1)/2}^2


    Someway to prove the formula:


    An^4+Bn^3+Cn^2+Dn+E=S_n

    put:
    n=1,2,3,4

    and solve the system of 4 equations, find A,B,C,D,E.
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  4. #4
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    Hello, fxs12!

    How about an inductive proof?


    \text{Prove that: }\;1^3  + 2^3  + 3^3  + \hdots + 100^3  \:=\: \left(\dfrac{100 \times 101}{2}\right)^2

    \text{We want to prove that: }\;1^3 + 2^3 + 3^3 + \hdots + n^3 \:=\:\left[\dfrac{n(n+1)}{2}\right]^2


    \text{Establish }S(1):\;\;1^2 \:=\:\left[\frac{(1)(2)}{2}\right]^2 . . . True.


    \text{Assume }S(k)\text{ is true: }\;1^3 + 2^3 + 3^3 + \hdots + k^3 \;=\;\left[\dfrac{k(k+1)}{2}\right]^2


    \text{Add }(k+1)^3\text{ to both sides:}

    . . 1^3 + 2^3 + 3^3 + \hdots + k^3 + (k+1)^3 \;=\;\left[\dfrac{k(k+1)}{2}\right]^2 + (k+1)^3 .[1]


    \text{The right side is: }\:\dfrac{k^2(k+1)^2}{4} + (k+1)^3 \;=\;(k+1)^2\bigg[\dfrac{k^2}{4} + k + 1\bigg]

    . . . . . . . . . . . . =\;(k+1)^2\bigg[\dfrac{k^2 + 4k + 4}{4}\bigg] \;=\;\dfrac{(k+1)^2(k+2)^2}{4}


    Therefore, [1] becomes: . 1^3 + 2^3 + 3^3 + \hdots + (k+1)^3 \;=\;\left[\dfrac{(k+1)(k+2)}{2}\right]^2

    The inductive proof is complete.

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