Proving the Sum of Cubes Formula using Mathematical Induction

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Homework Help Overview

The discussion revolves around proving the sum of cubes formula using mathematical induction, specifically the equation \(\sum^{n}_{i=1} i^{3} = \left[\frac{(n+1)\cdot n}{2}\right]^{2}\). Participants are exploring the steps involved in the induction process.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss proving the base case for \(n=1\) and assume the statement holds for an arbitrary \(n=p\). They express the need to show it holds for \(n=p+1\) and share their progress in manipulating the equation.

Discussion Status

Some participants have provided guidance on factoring and expanding terms to facilitate the induction step. There is an ongoing exploration of algebraic manipulation without a clear consensus on the next steps.

Contextual Notes

Participants are working under the constraints of mathematical induction and are focused on the algebraic aspects of the proof. There is an emphasis on expanding and simplifying expressions as part of the problem-solving process.

efekwulsemmay
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Nevermind, got the answer :D

Homework Statement


I need to prove using mathematical induction that:

[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]


Homework Equations



[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]


The Attempt at a Solution



Ok, I have proved that the statement is true when n=1 and have gone on to assume that it must be true for any arbitrary value of n. I chose n=p. Thus:

[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p ^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}[/tex]

Where i am stuck is that I know i need to show that if the statement is true when n=p then it must be true for n=p+1. I have gotten as far as:

[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}[/tex]
 
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[tex] \sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}[/tex]

Factor out a (p+1)^2, and see what you see.
 


efekwulsemmay said:

Homework Statement


I need to prove using mathematical induction that:

[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]


Homework Equations



[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]


The Attempt at a Solution



Ok, I have proved that the statement is true when n=1 and have gone on to assume that it must be true for any arbitrary value of n. I chose n=p. Thus:

[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p ^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}[/tex]

Where i am stuck is that I know i need to show that if the statement is true when n=p then it must be true for n=p+1. I have gotten as far as:

[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}[/tex]

have you tried just expanding it out and seeing what you get?
the induction step is simple, you got that - the rest is just algebra
 
Basically, you should expand, collect terms, and reduce. You should be able to do it that way.
 

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