proof by induction

**surjective** · February 21st 2010, 08:56 PM

Hello,

Could someone help me prooving the following:

If

$x+1 >= 0$

then

$(1+x)^n >= 1+nx$

for all natural numbers.

I hope that i have written everything correctly (this is my first time).

NB! How do i type in equations and stuff?

Thank you very much.

**o_O** · February 21st 2010, 09:29 PM

Please no double posting. On typing math symbols, see this thread for more: LaTex Tutorial

You can see the code that generates the image by clicking on it.

As for your problem, let's look at the inductive step. Assuming that $\color{red} (1+x)^k \geq 1 + kx$ , our goal is to prove that $(1+x)^{k+1} \geq 1 + (k+1)x$ .

Looking at the LHS, notice that: $(1+x)^{k+1} = {\color{red}(1+x)^k}(1+x)$

Now I highlighted the part in red for a reason .. and it shouldn't be too hard to show that this is bigger than $1 + (k+1)x$ .

**surjective** · February 22nd 2010, 12:30 AM

Hello again,

Sorry for the inconvenience. The thing is that what you mentioned before is exactly what I have already done . . . and then I get stuck. I can't seem to get the "great idea" (maybe because I havn't slept all night).

What I'm trying to do is to reach the conslusion (instead of showing the truthfulness of an inequality):

$(1+x)^{n+1} \geq 1+(n+1)x$

Now from calculus I know that:

$(1+x)^{n+1}= (1+x)^{n}(1+x)$

Using the inductive step we may write:

$(1+x)^{n}(1+x) \geq (1+nx)(1+x)$

This is where I get stuck.

Your help is greatly appreciated.

Thanks.

**Archie Meade** · February 22nd 2010, 03:21 AM

Originally Posted by surjective

Hello again,

Sorry for the inconvenience. The thing is that what you mentioned before is exactly what I have already done . . . and then I get stuck. I can't seem to get the "great idea" (maybe because I havn't slept all night).

What I'm trying to do is to reach the conslusion (instead of showing the truthfulness of an inequality):

$(1+x)^{n+1} \geq 1+(n+1)x$

Now from calculus I know that:

$(1+x)^{n+1}= (1+x)^{n}(1+x)$

Using the inductive step we may write:

$(1+x)^{n}(1+x) \geq (1+nx)(1+x)$

This is where I get stuck.

Your help is greatly appreciated.

Thanks.

Continuing on,

if $(1+x)^n\ \ge\ 1+nx,$ then the following must be true

$(1+x)^{n+1}\ \ge\ 1+(n+1)x$

Proof

$(1+x)^n(1+x)\ \ge\ (1+nx)(1+x)$ if the first statement is true

$(1+x)^n(1+x)\ \ge\ 1+x+nx+nx^2$

$(1+x)^{n+1}\ \ge\ [1+(n+1)x]+nx^2$

This means that if $(1+x)^n\ \ge\ 1+nx$

then $(1+x)^{n+1}\ \ge\ [1+(n+1)x]+nx^2$

and therefore

$(1+x)^{n+1}\ \ge\ 1+(n+1)x$ definately.

That's the inductive step.

**surjective** · February 22nd 2010, 04:18 AM

Hello,

Thank you very much. It seems that I was only one step away from completing the proof. Thank you all very much.

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