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Discrete Math 16 Online
OpenStudy (anonymous):

Prove each of the following for all n ≥ 1 by the Principle of Mathematical Induction. ∑[i=1,n]i^3=(n^2(n+1)^2/(4)=(∑[i=1,n,i])^2

OpenStudy (anonymous):

this is a well known sum too hard to write the proof here, but if you google it you will find many, including ones n youtube

OpenStudy (anonymous):

yeah I have looked and can not find anything could you try writing it out for me on here

OpenStudy (anonymous):

http://www.youtube.com/watch?v=ZtvCVgShG88

OpenStudy (anonymous):

case \(n=1\) as usual is trivial

OpenStudy (anonymous):

then it is algebra you get to assume \[1^2+2^3+3^3+...+k^3=\left(\frac{k(k+1)}{2}\right)^2\]

OpenStudy (anonymous):

then you need to show \[1^3+2^3+3^3+...+k^3+(k+1)^3=\left(\frac{(k+1)(k+2)}{2}\right)^2\]

OpenStudy (anonymous):

this boils down to showing \[\left(\frac{k(k+1)}{2}\right)^2+(k+1)^2=\left(\frac{(k+1)(k+2)}{2}\right)^2\]

OpenStudy (anonymous):

damn \[\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3=\left(\frac{(k+1)(k+2)}{2}\right)^2\]

OpenStudy (anonymous):

and that is a raft of elementary algebra

OpenStudy (anonymous):

so that is the answer

OpenStudy (loser66):

to the "prove" problem, you don't have answer, just the process how to get what they want you to prove. for this, you start at hypothesis step ( true when n =k) then add both sides of it by (k+1) ^2 , then, make some steps to show the final is the form of the case of (k+1)

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