Induction - have answer, need clarification

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Homework Help Overview

The discussion revolves around the mathematical induction proof for the formula of the sum of consecutive cubes, specifically the expression 1³ + 2³ + 3³ + ... + n³ = (n²(n + 1)²)/4. Participants are examining the algebraic steps involved in the proof and questioning the validity of certain transformations within the proof process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to follow the proof but expresses confusion regarding a specific algebraic manipulation. They question the transition from one expression to another, suggesting that the steps taken may not be valid. Other participants discuss the potential for factoring common terms instead of expanding, indicating different approaches to the problem.

Discussion Status

Some participants have provided clarifications regarding the algebraic manipulation, suggesting that the original poster may have misunderstood the factoring process. There appears to be a productive exchange of ideas, with some participants affirming the validity of the approach taken in the proof.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the depth of exploration into the proof's details. There is an emphasis on understanding the algebraic steps rather than simply arriving at a solution.

tangibleLime
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Homework Statement



Example 3 from this website:
http://www.themathpage.com/aprecalc/mathematical-induction.htm

Prove this formula for the sum of consecutive cubes:
1³ + 2³ + 3³ + . . . + n³ = (n²(n + 1)²)/4


Homework Equations





The Attempt at a Solution



I can understand and follow the entire thing up to a certain point. It seems like the algebra they start to do just doesn't follow. Where I get lost is at the location where they " To do that, add the next cube to S(k)". On the third step, they go from (k²(k + 1)² + 4(k + 1)³)/4 to ((k + 1)²[k² + 4(k + 1)])/4, which I am certain is nonsense. To back up my claims, I threw it into WolframAlpha and it said that those two expressions are not equal.

When I tried to do this myself, I just calculated (k+1)^3 and expanded it to (k^3+3k^2+3k+1), which is completely different from they they did. It's as if they just dropped the cube on the (k+1)?

Any clarification would be great.
 
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tangibleLime said:
On the third step, they go from (k²(k + 1)² + 4(k + 1)³)/4 to ((k + 1)²[k² + 4(k + 1)])/4, which I am certain is nonsense. To back up my claims, I threw it into WolframAlpha and it said that those two expressions are not equal.

When I tried to do this myself, I just calculated (k+1)^3 and expanded it to (k^3+3k^2+3k+1), which is completely different from they they did. It's as if they just dropped the cube on the (k+1)?

Any clarification would be great.

Um, I actually think that is true. You might have put it into Wolfram Alpha wrong. You can expand the cube, sure, but then you'll just want to factor it again. They just saved some time by taking advantage of common factors and not expanding.
 
Ooooh, okay, thanks! Just worked out the whole thing by hand and seems like it flows nicely now. Maybe I made a small mistake before.

Thanks again!
 
Like Dick said, they factorized out using common factors.

Say you wanted to find the factored form of (a+b)^3+2a(a+b)^2
rather than expanding everything out and then factorizing from there, which can be tedious and difficult to solve, they noticed that there is a common factor of (a+b)^2 and thus can factorize it as so: (a+b)^2\left((a+b)^1+2a\right)=(a+b)^2(3a+b).
 

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