Как свернуть формулу?

[AndreyXYZ]

$$\sum_{i=0}^{2k}{2k \choose {i}}{p}^{i}(1-p)^{2k-i}+\sum_{i=0}^{2k}{2k \choose {i}}{(-p)}^{i}(1-p)^{2k-i}=$$

$$=\sum_{j=0}^{k}{2k \choose {2j}}(2p^{2j})(1-p)^{2k-2j}=2\sum_{j=0}^{k}{2k \choose {2j}}p^{2j}(1-p)^{2k-2j}.$$