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

In summary, we are trying to prove the inequality (1+h)^n >= 1+nh+(n(n-1)/2)h^2 for any positive h and integer n>=0. We start by proving the base case P(0) is true, and then use the inductive hypothesis P(K) to show P(k+1) is true. However, there is a mistake in the calculation of P(k+1), as it should be n(n-1)/2 instead of n(n+1)/2. This can be seen by considering n=2 and comparing the results.
  • #1
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Homework Statement


Prove that for any positive h and any integer n[itex]\geq[/itex]0, (1+h)n[itex]\geq[/itex]1+nh+[itex]\frac{n(n+1)}{2}[/itex]h2.


Homework Equations


None.


The Attempt at a Solution


I proved that P(0) is true (1[itex]\geq[/itex]1). The rest of the proof goes as follows:

Assume K[itex]\in[/itex]Z (the set of integers) and P(K) is true.
Then (1+h)K[itex]\geq[/itex]1+Kh+[itex]\frac{K(K-1)}{2}[/itex]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+[itex]\frac{(K+1)(K+1-1)}{2}[/itex]h2.
 
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  • #2
You have that (1+h)K+1=(1+h)K(1+h) = P(K)(1+h), so see where that goes.
 
  • #3
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.
 
  • #4
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

[tex](1+h)^2=1+2h+h^2[/tex]

and not

[tex](1+h)^2=1+2h+3h^2[/tex]
 

What is the statement being proved by induction?

The statement being proved is (1+h)n ≥ 1+nh+ ½n(n+1)h2.

What is the purpose of using proof by induction?

The purpose of using proof by induction is to show that a statement holds for all natural numbers, by proving it for a base case and showing that if it holds for one number, it also holds for the next number.

What is the base case for this proof by induction?

The base case for this proof by induction is n = 1, where (1+h)1 ≥ 1+1h+ ½(1)(2)h2, which simplifies to 1+h ≥ 1+h, showing that the statement holds for n = 1.

What is the inductive hypothesis for this proof by induction?

The inductive hypothesis for this proof by induction is assuming that the statement holds for n = k, where k is any natural number.

What is the inductive step for this proof by induction?

The inductive step for this proof by induction is to show that if the statement holds for n = k, then it also holds for n = k+1. This is done by substituting k+1 into the original statement and using the inductive hypothesis to simplify and show that it is true.

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