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Here is the question: http://i.imgur.com/HnBo7.png
Base case n=1
1*(11)+2*(21)+3*(31)=1*(9*1^21)
1*(0)+2*(1)+3*(2)=1*(8)
8=8
How to do Inductive step for n=k and n=k+1?
 one year ago
 one year ago
Here is the question: http://i.imgur.com/HnBo7.png Base case n=1 1*(11)+2*(21)+3*(31)=1*(9*1^21) 1*(0)+2*(1)+3*(2)=1*(8) 8=8 How to do Inductive step for n=k and n=k+1?
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[\text{assume: }\sum_{1}^{3k}i(i1)=k(9k^21)\] \[\text{prove: }\sum_{1}^{3(k+1)}i(i1)=(k+1)(9(k+1)^21)\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[1(0)+2(1)+3(2)+...+k(k1)=k(9k^21)\] by adding (k+1)((k+1)1) to both sides we get \[1(0)+2(1)+...+k(k1)+[(k+1)(k+1)1]=k(9k^21)+[(k+1)(k+1)1\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
rewrite \[k(9k^21)+(k+1)[(k+1)1]~\text{ into }~(k+1)(9(k+1)^21)\]
 one year ago
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