Induction Proof

[Idontunderstand]

Use induction to show that (1+x)^n >= 1+n For all n element N where x>-1


(1+x)^(n+1)= (1+x)(1+x)^n
= (1+x)(1+xn)


I was wondering if somebody can explain how I am able to pull the n down to be next to the x? I also know that 1+x>0

 

[JeffM]

Use induction to show that (1+x)^n >= 1+n For all n element N where x>-1


(1+x)^(n+1)= (1+x)(1+x)^n
= (1+x)(1+xn) HUH? (1 + 4)^3 = 125 and (1 + 4)(1 + 2 * 4) = 5 * 9 = 45. What are you thinking?


I was wondering if somebody can explain how I am able to pull the n down to be next to the x? You can't because it is not true. I also know that 1+x>0 Correct IF YOU POSED THE PROBLEM CORRECTLY

I do not understand how much math you know. You asked a previous question involving proofs about transcendental functions of complex numbers. You also previously asked a riduclously easy question about the decimal number system. Now you are asking about a proof by induction. This seems to be all over the map.

What is your math background? What class are you taking? Do you understand how to do a proof by mathematical induction. In your class, do you define N to include 0 or not?

Formally, what are the two steps in a proof by induction?

Have you done the first step? If so, please show your work.

What is usually a good way to start the second step?

Edit: By the way, what exactly is the question asking? What you have said is to be proven cannot because it is false.

\(\displaystyle x = - 0.5\ and\ n = 2 \implies (1 + x)^n = (1 - 0.5)^2 = (0.5)^2 = 0.25 < 3 = 1 + 2.\)

 

[daon2]

This looks like the following "standard" inequality: \(\displaystyle (1+x)^n \ge 1+nx\), and OP probably wrote it down or typed it incorrectly.