Explanation

(a)
We have, $\ ^{4n}C_{0} + \ ^{4n}C_{2}x^{2} + \ ^{4n}C_{4}x^{4} + ... + \ ^{4n}C_{4n}x^{4n}$
$= \frac{1}{2}\left\lbrack (1 + x)^{4n} + (1 - x)^{4n} \right\rbrack$
On putting $x = 1$ and $x = i$, we get
$\ ^{4n}C_{0} + \ ^{4n}C_{2} + ... + \ ^{4n}C_{4n} = \frac{1}{2}\lbrack 2^{4n}\rbrack$ …(i)
and $\ ^{4n}C_{0} + \ ^{4n}C_{2} + ... + \ ^{4n}C_{4n} = \frac{1}{2}\left\lbrack (1 + i)^{4n} + (1 - i)^{4n} \right\rbrack$ …(ii)
On adding Eqs. (i) and (ii), we get
$2\left\lbrack \ ^{4n}C_{0} + \ ^{4n}C_{4} + ... + \ ^{4n}C_{4n} \right\rbrack = 2^{4n - 1} + \frac{1}{2}\lbrack{(1 + i)}^{4n} + {(1 - i)}^{4n}\rbrack$
Now,${(1 + i)}^{4n} + {(1 - i)}^{4n}$
$= \left\lbrack \sqrt{2}\left( \cos{\frac{\pi}{4} + i\sin\frac{\pi}{4}} \right) \right\rbrack^{4n} + \left\lbrack \sqrt{2}\left( \cos{\frac{\pi}{4} - i\sin\frac{\pi}{4}} \right) \right\rbrack^{4n}$
$= 2^{2n}\left( \cos{n\pi} + i\sin{n\pi} \right) + 2^{2n}\left( \cos{n\pi} - i\sin{n\pi} \right)$
$= 2^{2n + 1}\cos{n\pi} = 2^{2n + 1}{( - 1)}^{n}$
$\therefore 2\left\lbrack \ ^{4n}C_{0} + \ ^{4n}C_{4} + ... + \ ^{4n}C_{4n} \right\rbrack = 2^{4n - 1} + \frac{1}{2}2^{2n + 1}{( - 1)}^{n}$
$\Rightarrow \ ^{4n}C_{0} + \ ^{4n}C_{4} + ... + \ ^{4n}C_{4n} = 2^{4n - 1} + {( - 1)}^{n}2^{2n - 1}$