a^n +/- b^n

[Function]

Hello everyone

 

I'd just like to known if the formulas below are correct. Normally I wouldn't post such a short/maybe dumb question, but since Wolfram|Alpha won't generate these formulas, which are, according to me, somewhat obvious, I'd just like to know if there are correct.

 

[math]a^n-b^n=(a-b)\cdot \sum_{i=1}^n{\left(a^{n-i}\cdot b^{i-1}\right)}[/math]

 

[math]a^n+b^n=(a+b)\cdot \sum_{i=1}^n{\left((-1)^{i-1}\cdot a^{n-i}\cdot b^{i-1}\right)}[/math]

 

Thanks.

 

Function

[Sato]

Hello,

 

Inputting the expression "Simplify [//math:(a-b)*(sum (a^(n-i) * b^(i-1)), i=1 to n)//]" into WolframAlpha yields:

a^n + b^n for a != b

 

(as expected)

 

How did you derive these results? What was your thought process?

Edited May 10, 2014 by Sato

[Function]

Hello,

 

Inputting the expression "Simplify [//math:(a-b)*(sum (a^(n-i) * b^(i-1)), i=1 to n)//]" into WolframAlpha yields:

a^n + b^n for a != b

 

(as expected)

 

How did you derive these results? What was your thought process?

 

Blame it on my brains, which want to find a general formula for special expressions, like a^n +/- b^n

I first looked for n even, so I found:

[math]a^{2n}-b^{2n}=(a^n-b^n)(a^n+b^n)[/math]

 

And then for n odd, looked up what a^3+b^3, a^5+b^5, a^7+b^7, ... equals when facotized, and then I found the formulas in #1. So, I guess they only work if n is odd.

Edited May 10, 2014 by Function

[mathematic]

Hello everyone

 

I'd just like to known if the formulas below are correct. Normally I wouldn't post such a short/maybe dumb question, but since Wolfram|Alpha won't generate these formulas, which are, according to me, somewhat obvious, I'd just like to know if there are correct.

 

[math]a^n-b^n=(a-b)\cdot \sum_{i=1}^n{\left(a^{n-i}\cdot b^{i-1}\right)}[/math]

 

[math]a^n+b^n=(a+b)\cdot \sum_{i=1}^n{\left((-1)^{i-1}\cdot a^{n-i}\cdot b^{i-1}\right)}[/math]

 

Thanks.

 

Function

The first formula is true for all n. The second only for odd n.