# Why can we not expand $(a+b)^n$ directly when $n$ is a fractional or negative index?

Mathematics Asked on December 8, 2020

We know the binomial expansion of $$(1+x)^n$$ when $$n$$ is a fractional index or negative index. Why can we not expand $$(a+b)^n$$ directly when $$n$$ is a fractional or negative index? Instead of expanding directly we first take ‘$$a$$‘ common and write it as $$a^n(1+b/a)^n$$ and then we expand $$(1+b/a)^n$$ and multiply with $$a^n$$. It is written in my textbook that when $$n$$ is a fractional or negative index ‘$$a$$‘ must be equal to $$1$$ in $$(a+b)^n$$. But the result is same if we expand $$(a+b)^n$$ directly when $$n$$ is a fractional or negative index or first expand $$(1+b/a)^n$$ and then multiply $$a^n$$. I expanded $$(a+x)^n$$ by Taylor theorem also, the result is same. Then why can we not use binomial theorem to expand $$(a+b)^n$$ directly when $$n$$ is a fractional or negative index?

You can, in a way. The generalized binomial theorem affords a definition of $$binom{n}{k}$$, for $$ninBbb C$$ and integer $$kge0$$, such that$$(1+b/a)^n=sum_{kge0}binom{n}{k}(b/a)^k,$$or equivalently$$(a+b)^n=sum_kbinom{n}{k}a^{n-k}b^k,$$provided $$|a|>|b|$$. Note this modulus requirement prevents us exchanging $$a,,b$$ on the RHS, even though the LHS is symmetric. (Another issue with exchanging the exponents is that $$binom{n}{k},,binom{n}{n-k}$$ are in general no longer both defined, let alone equal, unless e make sure to write the definition of binomial coefficients in terms of Gamma functions rather than factorials & Pochhammer symbols.) Note also that our summing over all non-negative integers $$k$$ also holds when $$n$$ is a non-negative integer, because in that case any $$k>n$$ yields $$binom{n}{k}=0$$. This case also lets us drop the constraint $$|a|>|b|$$ altogether, so its presence when $$n$$ isn't a non-negative integer is very important.

Answered by J.G. on December 8, 2020

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