Proof by Induction - Divisibility Proofs

AI Thread Summary
The discussion centers on proving a divisibility statement by induction, specifically showing that a certain expression is divisible by 6. A participant expresses difficulty in their proof and seeks assistance in identifying errors. Another contributor points out incorrect usage of implication symbols and clarifies the transformation of the expression. The focus is on demonstrating that the rewritten expression maintains equality and can be factored to show divisibility by 6. The conversation emphasizes the importance of correctly manipulating algebraic expressions in proofs.
odolwa99
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Homework Statement



Q. Prove by induction that... (please see attachment).

Homework Equations





The Attempt at a Solution



The end result should be divisible by 6, but hasn't worked out for me. Can someone help me spot where I've gone wrong? Thank you.
 

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You are using implication symbols (==>) incorrectly. Most of the places you have them, you should be using "=".

As to your question,
7*7k + 4*4k + 1
= 7*7k + 7*4k + 7 - 3*4k - 6.

Can you show that the expression above has a factor of 6 in it?
 
May I ask how you solved for...
7*7k + 7*4k + 7 - 3*4k - 6
from
7*7k + 4*4k + 1?
 
odolwa99 said:
May I ask how you solved for...
7*7k + 7*4k + 7 - 3*4k - 6
from
7*7k + 4*4k + 1?
I didn't "solve" for it; I just rewrote the second expression in a different way. You should convince yourself that the two expressions are equal.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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...
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