Prove

**dhiab** · October 8th 2009, 08:28 PM

n is a naturel number

**HallsofIvy** · October 9th 2009, 04:57 AM

So the problem is to show that $2^{n+1}- 2^n+ 1= 0$ has at least on solution between 2 and $\frac{2n}{n+1}$ .

I'm sorry but that makes no sense. If n is "unknown" what does it mean to say that a solution is between 2 and a function of n?

**BobP** · October 11th 2009, 01:27 AM

?

The question is lacking a bit in definition, (is $n$ an integer and is it greater than 1 ?), but it does make sense.

(BTW, I'm still learning LaTex, can anyone tell me how to line up these n's with the rest of the text ?)

Assuming that $n$ is an integer greater than 1, it is saying, by way of example, that if

$n=2,$ the equation $x^3-2x^2+1=0$ has a root between $x=2$ and $x=4/3.$ It does, 1.618 (3d).

$n=3,$ the equation $x^4-2x^3+1=0$ has a root between $x=2$ and $x=6/4.$ It does, 1.839 (3d).

To show that the result is correct in general, define

$f(x)=x^{n+1}-2x^n+1.$

Then, $f(2)=2^{n+1}-2.2^n+1=1>0,$ and showing that $f(\frac{2n}{n+1})<0$ will be sufficient.

$f(\frac{2n}{n+1})=\frac{(2n)^{n+1}}{(n+1)^{n+1}}-2\frac{(2n)^n}{(n+1)^n}+1=1-\frac{2^{n+1}n^n}{(n+1)^{n+1}}=1-\frac{2^{n+1}/(n+1)}{(1+1/n)^{n}}.$

Without going into the analysis,the second term is always greater than 1, ( $n\geq2)$
(Check it out numerically for $n=2$ and $n=3$ and then use the fact that the numerator is an increasing function, while the denominator tends to e from below).

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