Prove P(n): "Proof by Induction for (1+h)n\geq1+nh+\frac{n(n+1)}{2}h2

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Homework Statement


Prove that for any positive h and any integer n\geq0, (1+h)n\geq1+nh+\frac{n(n+1)}{2}h2.


Homework Equations


None.


The Attempt at a Solution


I proved that P(0) is true (1\geq1). The rest of the proof goes as follows:

Assume K\inZ (the set of integers) and P(K) is true.
Then (1+h)K\geq1+Kh+\frac{K(K-1)}{2}h2.
Then (1+h)(K+1) = (1+h)K+(1+h)1...

I can't figure out how to relate that part to the final part of P(K+1), which is 1+(K+1)h+\frac{(K+1)(K+1-1)}{2}h2.
 
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You have that (1+h)K+1=(1+h)K(1+h) = P(K)(1+h), so see where that goes.
 
I'm working on the same problem, to show P(k+1) I set it up the same way

but then we can use our inductive hypothesis

(1+x)^(k+1) >= (1+kx+(1/2)*k(k-1)*x^2)(1+x)

My question is, I've wrestled with the algebra for a little while now and for some reason in my notes i had P(K) set to:

1+kx+(1/2)*k(k+1)*x^2

(where the 1 in k(k+1) is positive instead of negative) I think my professor did the problem with k+1 instead of k-1. But i thought from teh inductive hypothesis the term is k(k-1) NOT k(k+1) because k+1 is what we get from P(k+1) that is what we get from the substitution?

I know there will be left over terms but i keep getting k(k-1)/2 instead of k(k+1)/2.
 
stihl29 said:
I'm working on the same problem, to show P(k+1) I set it up the same way

but then we can use our inductive hypothesis

(1+x)^(k+1) >= (1+kx+(1/2)*k(k-1)*x^2)(1+x)

My question is, I've wrestled with the algebra for a little while now and for some reason in my notes i had P(K) set to:

1+kx+(1/2)*k(k+1)*x^2

(where the 1 in k(k+1) is positive instead of negative) I think my professor did the problem with k+1 instead of k-1. But i thought from teh inductive hypothesis the term is k(k-1) NOT k(k+1) because k+1 is what we get from P(k+1) that is what we get from the substitution?

I know there will be left over terms but i keep getting k(k-1)/2 instead of k(k+1)/2.

I believe you are correct. It has to be n(n-1) instead of n(n+1).

This is seen by picking n=2, then

(1+h)^2=1+2h+h^2

and not

(1+h)^2=1+2h+3h^2
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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