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As you know, we could write $(a+b+c)^2$ as $a^2+b^2+c^2+2ab+2ac+2bc$.
what about $(a+b+c+\cdots)^{1/2}$? is there any expansion for $(a+b+c+\cdots)^{1/2}$?
Any simplification or approximation of $(a+b+c+\cdots)^{1/2}$ could help me.
Thanks in advance

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1 Answer 1


If $|a|>|b|$ then \begin{align} (a+b)^{1/2} & = a^{1/2} + \frac12 a^{1/2-1} b + \frac{(1/2)(1/2-1)}{2\cdot1} a^{1/2-2} b^2 \\[12pt] & {}\quad{} + \frac{(1/2)(1/2-1)(1/2-2)}{3\cdot2\cdot1} a^{1/2-3} b^3 \\[12pt] & {}\quad{} + \frac{(1/2)(1/2-1)(1/2-2)(1/2-3)}{4\cdot3\cdot2\cdot1} a^{1/2-4} b^4 + \cdots \end{align}

But I'm not sure just how this should go with a sum of more than two terms, as in $(a+b+c)^{1/2}$. Notice that I said $|a|>|b|$, and notice the asymmetry in the expansion: $b$ is always raised only to integer powers; not so for $a$.

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You could extend this to sums with multiple terms as $(a+b+c+\cdots)^{1/2}=a^{1/2}+\frac12a^{1/2-1}(b+c+\cdots)+\frac{(1/2)(1/2-1)}{2‌​\cdot1}a^{1/2-2}(b+c+\cdots)^2+\cdots$, and then apply the usual binomial theorem to the terms $(b+c+\cdots)^n$ terms. Edit: As long as $|a|>|b+c+\cdots|$ of course. –  Rahul Aug 24 '13 at 16:22
@RahulNarain, if you factor out the $(b+c+\cdots)^{1/2}$, things get a little less nasty, at least if you're willing to accept sums in the denominators in the end. –  dfeuer Aug 24 '13 at 16:30

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