Induction Help (1 Viewer)

[Lukybear]

(1+x)^n>= 1+nx, x>0

 

[biopia]

(1+x)^n ≥ 1+nx
(1+x)^n - 1 - nx ≥ 0

I'll just go from the n=k part
Assume true for n=k
i.e. (1+x)^k - 1 - kx ≥ 0

Prove true for n=k+1
(1+x)^(k+1) - 1 - (k+1)x ≥ 0
= (1+x)(1+x)^k - 1 - kx - x
= (1+x)[(1+x)^k - 1 - kx] + kx^2
≥ 0 Since (1+x)^k - 1 - kx ≥ 0, and kx^2 ≥ 0

Therefore it is true for n=k+1 if it is true for n=k.

Since it is true for n=1, then it is true for n=2 and hence is true for n=3 and so on.

Hence it is true for all positive integers n.



Hope that helps =]

 

[Trebla]

 

[Lukybear]

Could you please explain the step of by assumption?

Also biopia, how did you get this step:= (1+x)[(1+x)^k - 1 - kx] + kx^2

 

[gurmies]

Assumption states that (1+x)^k > 1+kx. Therefore (1+x)(1+x)^k > (1+kx)(1+x); x>-1

 

[biopia]

In Trebla's answer, he has multiplied both sides of the assumption by (1+x) which will always be positive because x ≥ 0, so you can do it!

In my answer, it's a matter of deduction and calculation.
I have used the assumption [(1+x)^k - 1 - kx] and multiplied it by (1+x) also. So:
(1+x)[(1+x)^k - 1 - kx] = (1+x)(1+x)^k - (1+x) - kx(1+x)
= (1+x)(1+x)^k - 1 - x - kx - kx^2

This line is almost exactly the same as the second line of my proof stage, expect it has an extra [-kx^2]. To compensate for this, we add kx^2 to the assumption multiplied by (1+x).

i.e. (1+x)<i>[(1+x)^k - 1 - kx]</i> <b>+ kx^2</b>

Assumption is in italics, and the "compensation" is in bold.
This way is also correct because the assumption is ≥ 0, (1+x) ≥ 0 and kx^2 ≥ 0, as it is a squared number.

I hope that makes sense. If not, just say so and I'll try to rephrase it =]