# Extension of BinomialTheorem

1. May 10, 2012

### eddybob123

Thank you in advance, I need help proving or disproving this. In the binomial theorem, with a power (a+b)^n, I need to prove that a^n + b^n is greater than the rest, or in other words, (a+b)^n - (a^n + b^n).

2. May 10, 2012

### DonAntonio

Uh? I think you forgot to add some info and/or to write some symbols, as it seems to the question doesn't make sense as it is.

DonAntonio

3. May 10, 2012

### Matt Benesi

It depends on what you use for a,b, and n. In some cases, a^n + b^n will be greater than "the rest", in others, it won't.

4. May 11, 2012

### HallsofIvy

I think you mean that you want to prove that $(a+ b)^n- (a^n+ b^n)$ is positive. As Matt Benesi said, that depends upon what a and b are.

5. May 11, 2012

### dodo

If the statement is suspicious, the first thing is to try a few numerical examples, to try to find a counterexample. Here is a hint: 1+1=2. :)

P.S.: My understanding is that the OP tries to prove or disprove$$a^n + b^n > (a+b)^n - (a^n + b^n)$$

6. May 11, 2012

### DonAntonio

Too many assumptions: shall we let the OP to tell us what he meant, please?

DonAntonio

7. May 11, 2012

### eddybob123

That is correct. It can also be assumed that a and b do not equal 1.

8. May 11, 2012

### DonAntonio

Since the OP already wrote a post saying this is correct, this is the same as $$2(a^n+b^n)>(a+b)^n$$ which is greatly false, for example: for $\,\,a=1\,,\,b=2\,,\,n=3\,\,,\,\,or\,\,a=2\,,\,b=3\,,\,n=4\,\,$ , and infinite counterexamples more.

DonAntonio

9. May 11, 2012

### Matt Benesi

Question: can a and b be less than 1? Can they be negative?

Just some interesting cases: in all of these cases, x^n+y^n = "all the rest".

n=2 and a=b.

For n=3 we have $x=-y$, $x=y\times\left[2-\sqrt{3}\right]$, $x=y\times\left[2+\sqrt{3}\right]$.

Getting more complicated for n=4, just one example (out of 4):
$$x=\sqrt{2\,\sqrt{3}+3}\,y+\sqrt{3}\,y+y$$
or to preserve the format used above:
$$x=y\times\left[\sqrt{2\,\sqrt{3}+3}\,+\sqrt{3}\,+1\right]$$

Last edited: May 11, 2012
10. May 12, 2012

### eddybob123

a and b both must be integral and greater than 1. (a^n + b^n) must be greater than (a+b)^n - (a^n + b^n) by obvious reasoning. First of all, the two terms, a^n and b^n, contains the highest power of the binomial expansion. The next highest power,n-1, should be greater than a^n or b^n only when n is greater than the following coefficient

11. May 12, 2012

### DonAntonio

Either you don't understand mathematically what is going on here or else you're misunderstanding and/or misreading big time

the inequality you want/must prove.

You wrote above "(a^n + b^n) must be greater than (a+b)^n - (a^n + b^n) by obvious reasoning", which means Gauss knows what, but

this inequality is $\,\,a^n+b^n>(a+b)^n-(a^n+b^n)\Longleftrightarrow 2(a^n+b^n)>(a+b)^n\,\,$ , which already was show to be

false for lots and lots of options...Please do read and write carefully what you exactly want to achieve.

DonAntonio

12. May 12, 2012

### micromass

Please state exactly what you want to prove.