<s>[INST] Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$

Show this equation holds by squaring both sides and comparing terms up to $x^3$.

I wonder, how can I square the right hand side? 
 [/INST] Notice that
$$(a+b+c+d)^2=\underbrace{a^2+b^2+c^2+d^2}_{\text{the sum of square of all terms}}+\underbrace{2ab+2ac+2ad+2bc+2bd+2cd}_{\text{the sum of the double products of the terms taken 2 by 2 }}$$
so we find
$$\left(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 ..\right)^2=\underbrace{ 1^2+(2\times 1\times\frac12 x)}_{=1+x}+\underbrace{(\frac12 x)^2-2\times \frac18x^2}_{=0}\\\underbrace{-2\times \frac12x\frac18x^2+2\times\frac1{16}x^3}_{=0}+\cdots$$
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