Question

Bernoulli's Inequality - Prove that for all a in R and all n in N if, 1 + a >0, then (1+a)^n greater than or equal to 1 +na

Answer 1

For all \[a \in \mathbb{R} \] and all n\[\in \mathbb{N}\], if \[1+ a > 0\], then \[(1+a)^n \geq 1+na \]

Answer 2

What I got so far is that a is an element of real numbers and n belongs to natural numbers.

Answer 3

IT's like for this to work, my a has to be a positive number... but this is kind of a lame attempt at proving this statement.

Answer 4

@LastDayWork @wio

Answer 5

do I use induction or something?

Answer 6

doesn't really say :/...

Answer 7

Well, you could since \(n\in \mathbb N\)

Answer 8

You may need to use the binomial theorem.

Answer 9

Yes, agree with @wio

Answer 10

\[
(1+a)^n= \sum_{k=0}^{n}\binom nka^k
\]

Answer 11

I was reading about the binonmial theorem... but since the example didn't match up I thought that was pointless... dang...that makes sense since the theorem states that for any a, b, element in R and for any n in element N (a+b)^n

Answer 12

I get overwhelmed when lectures are all over the place and there's homework that's super behind and due in 48 hours T_T . if it's like the weekend like when I read about set theories more, I got what it was talking about.

Answer 13

Yep, we are all so sorry for you.

Answer 14

oh I see it... -___- (1+a)^n

Answer 15

then apply the binomial theorem to that piece of info.

Answer 16

dude I got my revised paper rejected again. UGHHHHHHHHH! At least I'm not the only one with that problem

Answer 17

Wow, really?

Answer 18

eeyup.... that's another reason why I don't want to do this s****. It's just going to be handed back to me without a score. -_-.

Answer 19

even if it is correct BAM!