Hey everyone, I have a question on some proof from an old exam. Please excuse my Latex, I have not yet mastered this.(adsbygoogle = window.adsbygoogle || []).push({});

Thanks for the help.

Tal

[tex[/tex]1. The problem statement, all variables and given/known data

for any [tex]1>t>-1 and p \geq 1[/tex] prove that [tex] (1+t)^p + (1-t)^p \geq 2 [/tex]

Thats all the info the question has. I am assuming p is natural.

2. Relevant equations

3. The attempt at a solution

Proof by induction. it is easy to see that p=1 fullfills what is required. So we shall assum for n and show that this invokes n+1.

[tex] (1+t)^{n+1} + (1-t)^{n+1} =\sum^{i=0}_{n+1}(\stackrel{n+1}{i})t^i + \sum^{i=0}_{n+1}(\stackrel{n+1}{i}) t^i [/tex]

[tex] = \sum^{i=0}_{n+1}(\stackrel{n+1}{i})t^i + \sum^{i=0}_{(n+1)/2} (\stackrel{n+1}{(2i)}) t^{2i} - \sum^{i=1}_{(n+1)/2} (\stackrel{n+1}{(2i +1)}) t^{2i+1} = 2 + \sum^{i=1}_{n+1}(\stackrel{n+1}{i}) t^{i} + \sum^{i=1}_{(n+1)/2} (\stackrel{n+1}{(2i)}) t^{2i} - \sum^{i=1}_{(n+1)/2} (\stackrel{n+1}{(2i+1)}) t^{2i+1} [/tex]

Since 1>t>-1 its succesive powers are smaller then the previous ones. Then we can see that the two partial sums of even and odd powers, the even powers are always larger then the odd powers and there sum is positive. The full sum is also positive for any t>0. but if we look at the original problem then we can see that if t < 1 the equation remains the same and so we always have a sum of 2 + something positive.

I wonder if this is correct.

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# Homework Help: Proof by induction series help

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