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 \( A^x B^y \). 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 \( |x| < 1 \): \[ (1+x)^z = \sum_{k=0}^\infty \binom{z}{k} x^k = 1 + zx + \frac{z(z-1)}{2}x^2 + \frac{z(z-1)(z-2)}{6}x^3 + \cdots. \] 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 \( (a+b)^{-5} = b^{-5} - 5a b^{-6} + 15a^2 b^{-7} - 35a^3 b^{-8} + \cdots \), if \( |a/b| < 1 \). Otherwise, we interchange the roles of \( a, b \). Exercise: if \( |a/b| = 1 \), then what happens?

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 ;)

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TopNewestIf 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 \( A^x B^y \). 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 \( |x| < 1 \): \[ (1+x)^z = \sum_{k=0}^\infty \binom{z}{k} x^k = 1 + zx + \frac{z(z-1)}{2}x^2 + \frac{z(z-1)(z-2)}{6}x^3 + \cdots. \] 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 \( (a+b)^{-5} = b^{-5} - 5a b^{-6} + 15a^2 b^{-7} - 35a^3 b^{-8} + \cdots \), if \( |a/b| < 1 \). Otherwise, we interchange the roles of \( a, b \). Exercise: if \( |a/b| = 1 \), then what happens?

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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}

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