Function Posted May 10, 2014 Share Posted May 10, 2014 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 Link to comment Share on other sites More sharing options...
Sato Posted May 10, 2014 Share Posted May 10, 2014 (edited) 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 Link to comment Share on other sites More sharing options...
Function Posted May 10, 2014 Author Share Posted May 10, 2014 (edited) 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 Link to comment Share on other sites More sharing options...
mathematic Posted May 10, 2014 Share Posted May 10, 2014 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. Link to comment Share on other sites More sharing options...
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