# Two variabe induction question

1. Nov 28, 2008

### transgalactic

n1 is the smallest whole number for which this inequality works :

(1+x)^n >1 +n*x + n*x^2

also i am given that x>0

find n1

and prove this inequality for every n=>n1 by induction.

the base case:

(1+x)^n1 >1+n1*x + n1*x^2

i think its correct because i was told that this inequality works for n1.

n=k step we presume that this equation is true :

equation 1: (1+x)^k >1 +k*x + k*x^2

n=k+1 step we need to prove this equation:
equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

now i need to multiply equation1 by sum thing
and
do
if a<b<c
then a<c

how to do this thing in this case?

Last edited: Nov 28, 2008
2. Nov 28, 2008

### transgalactic

i multiplied equation 1 by (1+x)
(1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)

i tried to do this
but its not working
(1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)>(1 +k*x + k*x^2)

(1+x)^(k+1) >(1 +k*x + k*x^2)

??

3. Nov 28, 2008

### e(ho0n3

First, have you found n1? I suggest using the strong form of induction: Assume the inequality is valid for all whole numbers less n.

4. Nov 28, 2008

### transgalactic

what is the general way of finding n1?

whats strong form of induction?

i know
n=k
n=k+1

??

5. Nov 28, 2008

### e(ho0n3

I don't know of a general way. I would try n1 = 1, 2, 3, etc. and see which one works.

Instead of assuming n = k and proving n = k + 1, you assume the proposition is true for all k < n and then prove the proposition for n.