General

[quote=AnasF]I was thinking about that, but it seemed like a painful/inelegant way to prove it. :x
I prefer algebraic proofs to series. Personal preference. x_x

@vaelietta: Check my proof in the first post. See if it matches anything in your class, or seems correct. I'm interested to see if my randomness is actually correct. [/quote]

I don't have classes until tomorrow, but it's funny. I just found the exact same problem online. [url=http://mathforum.org/library/drmath/view/51613.html]lol[/url]

[quote=WeffTooFat]Kinda of a stupid way to use induction and waste IMO, but w/e. It sounds like the problem just hands the answer out to you. here's the proof in a nut shell:

Assume for n, (1+x)^n >= 1 + nx.

Use binomial theorem: 1 + nC1 x + nC2 x^2 + ... + nCn x^n = 1 + nx + nC2 x^2 + ... + x^n >= 1 + nx

nC2 x^2 + ... + x^n >= 0

True for n+1?

Use binomial theorem: 1 + (n+1)C1 x + (n+)C2 x^2 + ... + (n+1)C(n+1) x^(n+1) = 1 + (n+1)x + (n+1)C2 x^2 + ... + x^(n+1) >= 1 + (n+1)x
(n+1)C2 x^2 + ... + x^(n+1) >= 0

Too see this more clearly, let k + n+1:
kC2 x^2 + ... + x^k >= 0

yes, it is true!

There's your proof.[/quote]
I was thinking about that, but it seemed like a painful/inelegant way to prove it. :x
I prefer algebraic proofs to series. Personal preference. x_x

@vaelietta: Check my proof in the first post. See if it matches anything in your class, or seems correct. I'm interested to see if my randomness is actually correct.

Lolol I see NX.

Am I 12 now?

[quote=LeechLess]Just curious, what math are you in ? Algebra II ?[/quote]

I'm taking discrete math and calculus. Calculus is easy mode because my professor teaches it with concrete math. My discrete math professor gets into the theories and proofs and aawwwh my god. So I'm basically learning from the book and this was a homework question.

Just curious, what math are you in ? Algebra II ?

[quote=littlezz]hmm oh i see Sorry I dunno how then.[/quote]

It's cool bro. I don't expect a lot of people to be taking discrete math. I'm lucky to get any answer.

[quote=littlezz]Are we trying to prove x > than -1?

Bleh I didn't get the answer, but maybe it might still help?

Hmm my guess is to take the derivative of both sides, so

n (1+x)^(n-1) => n

Divide by n

(x+1)^(n-1) => 1

Take the (n-1)th root of both sides.

X+1 => 1 because 1 times itself no matter how many times is 1

X => 0 .... I got close D:

Sorry if that wasn't even your question >.<[/quote]
a) It asks to use induction.
b) It asks to prove (1+x)^n => 1+nx

Induction means you assume it's true for n, then prove that it works for n+1, using that assumption.

Assume, (1+x)^n => 1+nx
Now, test for n+1
(1+x)^[n+1] => 1+(n+1)x
(1+x)(1+x)^n => 1 + nx + x
(1+x) => [1+nx+x]/(1+x)^n
(1+x) => [1+nx]/(1+x)^n + x/(1+x)^n

Now, since you're assuming that (1+x)^n => 1+nx, you know that [1+nx]/(1+x)^n =< 1.
Thus, show that (1+x) - x/(1+x)^n => 1. Idk how to explain why I got to this step, but write it out on paper with [1+nx]/(1+x)^n = a, where a =< 1, and it should make sense to you.

(1+x) - x/(1+x)^n => 1
(1+x)^[n+1] - x => (1+x)^n
(1+x)(1+x)^n -x - (1+x)^n => 0
x(1+x)^n => x
(1+x)^n => 1
You can show that for all natural numbers n, and x > -1, this holds.

Edit: Just noticed, this is probably a very convoluted proof. However, I [b]just[/b] woke up, and I've never studied induction before, so meh.