Help with Mathematical Induction

AI Thread Summary
The discussion focuses on proving by mathematical induction that the expression (2^(n+1) + 9(13^n)) is divisible by 11 for all positive integers. Participants emphasize the importance of establishing a base case and assuming the proposition holds for n = k to demonstrate it for n = k+1. There is a suggestion to utilize modular arithmetic to simplify the proof process. The conversation includes some confusion about identities, with one participant clarifying their intent to assist. Overall, the key steps in the induction process are highlighted as essential for solving the problem.
SeattleScoute
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Homework Statement



Prove by matematical induction that (2^(n+1)+9(13^n)) divides by by 11 for all positive intergers


Homework Equations





The Attempt at a Solution



I really have no idea where to start...
 
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The problem outright tells you a place to start!
 
Is P1 true? If so, then if I say Pk is true is Pk+1 also true?
 
jegues said:
Is P1 true? If so, then if I say Pk is true is Pk+1 also true?
Are you also SeattleScoute?

That's basically what is needed. Establish a base case. Assume the proposition is true for n = k. Show that P(k) being true implies that P(k+1) is also true.
 
Are you also SeattleScoute?

No I'm not, I thought I'd help :S
 
you may use modular arithmetic to lighten your job .
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

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
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Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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