Ask your own question, for FREE!
Discrete Math 88 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)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Thayes: Que estu00e1s haciendo en este momento
20 seconds ago 14 Replies 2 Medals
Thayes: What u think?
25 minutes ago 6 Replies 0 Medals
HotPockets14: Why does it tell me i joined in Feb 2023 when I joined in 2020
45 minutes ago 5 Replies 2 Medals
abound: @notmeta is harassing me
40 minutes ago 70 Replies 2 Medals
TheJag18: Who has taken U.S. Government and Is good at it?
52 minutes ago 3 Replies 2 Medals
Rileyy: I need help
1 hour ago 11 Replies 1 Medal
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!