Binomial expansion going wrong! I am trying to expand
$$1 - \left(1-\frac{0.511}{v}\right)^{0.5}\left(1+\frac{0.511}{v}\right)^{0.5} $$
for large $v$, however using a binomial expansion on the two terms I get the expression 
$$\frac{0.511^2}{4v^2}$$
which is exactly half the correct answer.
Can anyone explain what is going wrong here?

Context from a Comment to an Answer:

My working is:
$$1 - \left(1-\frac{0.511}v\right)^{0.5}\left(1+\frac{0.511}v\right)^{0.5}$$ 
  I use the first two terms of the expansion
  $= 1 - \left(1-\frac{0.511}{2v}\right)\left(1+\frac{0.511}{2v}\right) = \frac{0.511^2}{4v^2}$ (**)
However this is not right because if I take a different route to the answer and spot that what is square rooted is the difference of 2 squares I obtain:
$$1 - \left(1-\frac{0.511^2}{v^2}\right)^{0.5}$$ 
  I use the first two terms of the expansion
  $= 1 - 1 + \frac{0.511^2}{2v^2} = \frac{0.511^2}{2v^2}$
Here lies the confusion, I am sure something is going wrong here (**) as before that I obtain the same result from both methods when I set v=100

 A: Not enough terms have been included. An answer is being computed to the second order using information only to the first order. The second order terms need to be included in each factor to get the correct second order term in the result:
$$
\begin{align}
&\left(1-\frac{0.511}{v}\right)^{0.5}\\
&=1+\overbrace{\ \left(\frac{0.5}1\right)\ }^{\large\binom{0.5}{1}}\left(-\frac{0.511}{v}\right)+\overbrace{\left(\frac{0.5}1\right)\left(\frac{-0.5}2\right)}^{\large\binom{0.5}{2}}\left(-\frac{0.511}{v}\right)^2+O\!\left(\frac1{v^3}\right)\\[12pt]
&=1-\frac{0.511}{2v}-\frac{0.511^2}{8v^2}+O\!\left(\frac1{v^3}\right)\tag{1}
\end{align}
$$
and
$$
\begin{align}
&\left(1+\frac{0.511}{v}\right)^{0.5}\\
&=1+\overbrace{\ \left(\frac{0.5}1\right)\ }^{\large\binom{0.5}{1}}\left(\frac{0.511}{v}\right)+\overbrace{\left(\frac{0.5}1\right)\left(\frac{-0.5}2\right)}^{\large\binom{0.5}{2}}\left(\frac{0.511}{v}\right)^2+O\!\left(\frac1{v^3}\right)\\[12pt]
&=1+\frac{0.511}{2v}-\frac{0.511^2}{8v^2}+O\!\left(\frac1{v^3}\right)\tag{2}
\end{align}
$$
Multiply $(1)$ and $(2)$ together:
$$
\begin{align}
&\left(1-\frac{0.511}{2v}-\color{#C00}{\frac{0.511^2}{8v^2}}+O\!\left(\frac1{v^3}\right)\right)\left(1+\frac{0.511}{2v}-\color{#090}{\frac{0.511^2}{8v^2}}+O\!\left(\frac1{v^3}\right)\right)\\
&=1-\frac{0.511}{2v}+\frac{0.511}{2v}-\frac{0.511^2}{4v^2}-\color{#C00}{\frac{0.511^2}{8v^2}}-\color{#090}{\frac{0.511^2}{8v^2}}+O\!\left(\frac1{v^3}\right)\\
&=1-\frac{0.511^2}{2v^2}+O\!\left(\frac1{v^3}\right)\tag{3}
\end{align}
$$
The second order terms are significant. Their omission in the computation caused the error.
A: hint
$$(1+a)^\alpha(1-a)^\alpha=(1-a^2)^\alpha$$
$$(1-b)^\alpha=1-\alpha\cdot b $$ for very small b.
