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tux

  • 3 years ago

Here is the question: http://i.imgur.com/HnBo7.png Base case n=1 1*(1-1)+2*(2-1)+3*(3-1)=1*(9*1^2-1) 1*(0)+2*(1)+3*(2)=1*(8) 8=8 How to do Inductive step for n=k and n=k+1?

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  1. amistre64
    • 3 years ago
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    if the basis works, then assume it works for any k, then adjust it by k+1 and try to get it back to a form of "k" to compare with

  2. amistre64
    • 3 years ago
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    \[\text{assume: }\sum_{1}^{3k}i(i-1)=k(9k^2-1)\] \[\text{prove: }\sum_{1}^{3(k+1)}i(i-1)=(k+1)(9(k+1)^2-1)\]

  3. amistre64
    • 3 years ago
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    \[1(0)+2(1)+3(2)+...+k(k-1)=k(9k^2-1)\] by adding (k+1)((k+1)-1) to both sides we get \[1(0)+2(1)+...+k(k-1)+[(k+1)(k+1)-1]=k(9k^2-1)+[(k+1)(k+1)-1\]

  4. amistre64
    • 3 years ago
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    rewrite \[k(9k^2-1)+(k+1)[(k+1)-1]~\text{ into }~(k+1)(9(k+1)^2-1)\]

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