Induction - have answer, need clarification

  • Thread starter Thread starter tangibleLime
  • Start date Start date
  • Tags Tags
    Induction
AI Thread Summary
The discussion centers on proving the formula for the sum of consecutive cubes using mathematical induction. A participant expresses confusion over a specific algebraic step in the proof, questioning the validity of transitioning from one expression to another. They initially believe the algebra is incorrect, as verified by WolframAlpha, but later receive clarification that the authors used factoring techniques to simplify the expression without fully expanding it. This approach ultimately leads to a better understanding of the proof's flow. The conversation highlights the importance of recognizing common factors in algebraic manipulations.
tangibleLime
Messages
71
Reaction score
0

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.
 
Physics news on Phys.org
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).
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top