## Proof by Induction

1. The problem statement, all variables and given/known data

$$(1+x)^{k}$$$$\geq$$1+kx

2. Relevant equations

3. The attempt at a solution
I want to show for P(k+1)
(1+x)^(k+1)$$\geq$$1+kx+x
(1+x)^k*(1+x)$$\geq$$1+kx+x

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 Quote by kathrynag (1+x)^k*(1+x)$$\geq$$1+kx+x
Use this:

 $$(1+x)^{k}$$$$\geq$$1+kx

 I don't really understand how I use that...

## Proof by Induction

You want to use an inequality involving $$(1 + x)^k$$ to derive an inequality involving $$(1+x)^k (1+x)$$, which is a multiple of it.

An inequality still holds if you multiply both sides by a positive number. If you multiply both sides of an inequality by a negative number, then you have to flip the sign from $$\le$$ to $$\ge$$ or vice versa.

 Ok, so: (1+x)^k(1+x)>(1+kx)(1+x) >(1+2kx+x)
 Right idea, but there is an error in your expansion. Does the question give any restrictions on x? Is what you wrote still true of 1 + x < 0?
 No restrictions. 1+kx+x+kx^2
 When k = 3 and x = -4, $$(1 + x)^k = (-3)^3 = -27 < -11 = 1 - 12 = 1 + kx$$ which makes the statement false.
 So, we have to assume 1+x>0
 so, x>-1

Mentor
 Quote by kathrynag So, we have to assume 1+x>0
I believe x has to be strictly within 1 unit of 1; i.e., |1 + x| < 1, which means that 0 < x < 2.

 Quote by kathrynag No restrictions. 1+kx+x+kx^2
Once I get here I'm unsure where to go

 Quote by kathrynag Once I get here I'm unsure where to go
Look at your first post; you need to show that that is $$\ge 1 + kx + x$$.