The value of $a_0 ^2-a_1 ^2+a_2 ^2....a_{2n} ^2$ is

$(1+x)^{2n}=1+_{ }^{ 2n }{ { C }_{ 1 }x+_{ }^{ 2n }{ { C }_{ 2 }{ x }^{ 2 }+....+_{ }^{ 2n }{ { C }_{ 2n }{ x }^{ 2n } } } }$

or,

$(1+x)^{2n}=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$ ...$(i)$

Similarly,

$(1-x)^{2n}=a_0-a_1x+a_2x^2+...+a_{2n}x^{2n}$ ....$(ii)$

As, $a_0=a_{2n}$, $a_1=a_{2n-1},...$and so on. So, (i) and (ii) can be written as:

$(1-x)^{2n}=a_{2n}-a_{2n-1}x+a_{2n-2}x^2+...+a_{0}x^{2n}$

$(1+x)^{2n}=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$

So, required answer is coefficient of $x^{2n}$ in $(1+x)^{2n}.(1-x)^{2n}=$ coefficient of $x^{2n}$ in $(1-x^2)^{2n}$

${T}_{r+1}=_{ }^{2n}{C}_r(-x^2)^r$

$\quad\quad=_{ }^{2n}{{C}_{r}}(-1)^r(x)^{2r}$

So, we need, $r=n$. Hence, the answer is $(-1)^n.a_n$

For $n-even$, the answer is option $A$, i.e; $a_n$