# Approximate $\sqrt{7}$ using binomial theorem [duplicate]

How does one deduce the approximation of $$\sqrt{7}$$ to be $$\frac{10837}{4096}$$ by taking $$x = \frac{1}{64}$$ in the expansion of $$\sqrt{1-x}$$?

How should you approach such a question? I assume the first step would be to expand $$\sqrt{1-x}$$ which can only be done through binomial theorem (afaik).

That gives $$\sqrt{1-x} = 1 - \frac{1}{2}(x) - \frac{1}{8}(x^2)$$ + ... and so on.

How do you continue? I can't seem to figure out how taking $$x = \frac{1}{64}$$ accomplishes anything.

• Well, $$\sqrt7=\frac83\sqrt{1-\frac1{64}}=\frac83\left(1-\frac12\frac1{64}-\frac18\left(\frac1{64}\right)^2+\ldots\right)$$ hence $$\sqrt7\approx\frac{8\cdot(8\cdot(64)^2-4\cdot64-1)}{3\cdot8\cdot(64)^2}=\cdots$$
– Did
Sep 29, 2018 at 9:14
• Forgive my ignorance but, how did you get $\sqrt7 = \frac{8}{3}\sqrt{1-\frac{1}{64}}$? Sep 29, 2018 at 9:18
• @gimusi. They don't give Field medals here ? I am really disappointed to hear that. Cheers. Sep 29, 2018 at 9:33
• @gimusi The friendliness in this stack exchange is refreshing. Thanks! Sep 29, 2018 at 9:40
• @Sam. You well noticed this key aspect of the site ! Sep 29, 2018 at 9:42

$$\sqrt{7}=\sqrt{9\cdot \frac79}=3\sqrt{\frac79}=3\sqrt{1-\frac29}$$
$$\sqrt{1-\frac1{64}}=\sqrt{\frac{63}{64}}=\frac38\sqrt 7$$