algebra precalculus - Find the sum $sum_{j=0}^{n}binom{4n}{4j}$

Find the sum of the series
$$\binom{4n}{0}+\binom{4n}{4}+\binom{4n}{8}+\ldots+\binom{4n}{n}=\sum_{j=0}^{n}\binom{4n}{4j}.$$

My approach is to consider $(1+x)^{4n} = \sum_{j=0}^{4n}\binom{4n}{j}x^j.$

How to proceed further now? Please help on this since I am quite clueless. Thanks a lot .