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Math
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a_{i-1}
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Comments
I myself derived the Binomial Theorem before it was handed to me... But I've only done the expansion for (a+b)^n, not -n... The only difference will be, then, I guess, all of the terms will be flipped.
Note that simply negating all the exponents for expansion of your problem wouldn't get the job done.
This must be the right expansion. I know it inside out, after all ;)
If by "expansion" you mean the series expansion, then none of the answers posted here so far is correct, because this must be expressed as a sum of terms of the form AxBy. Instead, the answers given so far are all reciprocals of polynomial expansions (and at least one of them isn't even mathematically correct).
The series expansion for a positive integer n can be derived in a number of ways, but we can also just look at the generalized binomial series formula, valid for any complex number z and ∣x∣<1: (1+x)z=k=0∑∞(kz)xk=1+zx+2z(z−1)x2+6z(z−1)(z−2)x3+⋯. Assuming 0<A<B are real, for example, then we can simply write (A+B)−n=B−n((A/B)+1)−n and apply the theorem. But we can also easily apply it for other cases, including complex values.
For a simple example, let n=5. Then (a+b)−5=b−5−5ab−6+15a2b−7−35a3b−8+⋯, if ∣a/b∣<1. Otherwise, we interchange the roles of a,b. Exercise: if ∣a/b∣=1, then what happens?
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
I myself derived the Binomial Theorem before it was handed to me... But I've only done the expansion for (a+b)^n, not -n... The only difference will be, then, I guess, all of the terms will be flipped.
(a+b)^n=a^n+...+n!/(k!(n-k)!(b^n-k)+...+b^k (a+b)^-n=1/(a+b)^n=1/{a^n+...+n!/(k!(n-k)!)(b^(n-k))+...+b^k}
Note that simply negating all the exponents for expansion of your problem wouldn't get the job done. This must be the right expansion. I know it inside out, after all ;)
In your terms:
(A+B)^-N=1/{A^N+...+N!/(k!(N-k)!)(N^(N-k))+...+B^k}
If by "expansion" you mean the series expansion, then none of the answers posted here so far is correct, because this must be expressed as a sum of terms of the form AxBy. Instead, the answers given so far are all reciprocals of polynomial expansions (and at least one of them isn't even mathematically correct).
The series expansion for a positive integer n can be derived in a number of ways, but we can also just look at the generalized binomial series formula, valid for any complex number z and ∣x∣<1: (1+x)z=k=0∑∞(kz)xk=1+zx+2z(z−1)x2+6z(z−1)(z−2)x3+⋯. Assuming 0<A<B are real, for example, then we can simply write (A+B)−n=B−n((A/B)+1)−n and apply the theorem. But we can also easily apply it for other cases, including complex values.
For a simple example, let n=5. Then (a+b)−5=b−5−5ab−6+15a2b−7−35a3b−8+⋯, if ∣a/b∣<1. Otherwise, we interchange the roles of a,b. Exercise: if ∣a/b∣=1, then what happens?