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Thread: Proof by Induction

  1. #1
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    Proof by Induction

    I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

    Prove $\displaystyle 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2$

    I'm basically looking for a shove in the right direction, as my induction step:

    $\displaystyle P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ??? $

    I'm unsure here since the RHS can be: $\displaystyle (1 + 2 + ... + k + [k+1])^2 $ or $\displaystyle (2 + 3 + ... + 2k)^2 $

    which is where my confusion begins. Thoughts?
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  2. #2
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    Quote Originally Posted by flabbergastedman View Post
    I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

    Prove $\displaystyle 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2$

    I'm basically looking for a shove in the right direction, as my induction step:

    $\displaystyle P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ??? $

    I'm unsure here since the RHS can be: $\displaystyle (1 + 2 + ... + k + [k+1])^2 $ or $\displaystyle (2 + 3 + ... + 2k)^2 $

    which is where my confusion begins. Thoughts?
    hi flabbergastedman,

    If $\displaystyle 1^3+2^3+3^3+....+n^3=(1+2+3+....+n)^2$

    then $\displaystyle 1^3+2^3+3^3+....+n^3+(n+1)^3$ should $\displaystyle =[1+2+3+...+n+(n+1)]^2$

    Hence

    $\displaystyle (1+2+3+....+n+(n+1))^2$$\displaystyle =(1+2+3+...+n)^2+(n+1)[1+2+3+...+n+(n+1)]+(1+2+3+...+n)(n+1)$

    $\displaystyle =(1+2+3+...+n)^2+2(n+1)(1+2+3+...+n)+(n+1)^2$

    $\displaystyle =(1+2+3+...+n)^2+\frac{2(n+1)(n(n+1)}{2}+(n+1)^2$

    since $\displaystyle (1+2+3+...+n)=\frac{n(n+1)}{2}$

    giving

    $\displaystyle [1+2+3+...+n+(n+1)]^2=(1+2+3+...+n)^2+n(n+1)^2+(n+1)^2$

    $\displaystyle =(1+2+3+...+n)^2+(n+1)^2(n+1)=(1+2+3+...+n)^2+(n+1 )^3$

    Proven
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  3. #3
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    Quote Originally Posted by flabbergastedman View Post
    I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

    Prove $\displaystyle 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2$

    I'm basically looking for a shove in the right direction, as my induction step:

    $\displaystyle P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ??? $

    I'm unsure here since the RHS can be: $\displaystyle (1 + 2 + ... + k + [k+1])^2 $ or $\displaystyle (2 + 3 + ... + 2k)^2 $

    which is where my confusion begins. Thoughts?

    $\displaystyle 1^3+2^3+\ldots+k^3+(k+1)^3=\left(1+2+\ldots+k\righ t)^2+(k+1)^3$ , using the inductive hypothesis, and now use that $\displaystyle 1+2+\ldots +n=\frac{n(n+1)}{2}$

    Tonio
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