Câu hỏi 29Tìm $$ \sum_{k=0}^{49}(-1)^k\binom{99}{2k}, $$ trong đó $$ \binom{n}{j}=\frac{n!}{j!(n-j)!}. $$$-2^{50}$$-2^{49}$$0$$2^{49}$$2^{50}$