Any simplification or approximation of sqrt

Question

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

Answer 1

http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem

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 $a$.