Câu hỏi 35
(MON 6) Chứng minh đồng nhất thức
$$\sum_{k=0}^n\binom nk\left(\tan\frac x2\right)^{2k}\left[1+2^k\frac1{(1-\tan^2(x/2))^k}\right]=\sec^{2n}\frac x2+\sec^n x.$$
$$\sum_{k=0}^n\binom nk\left(\tan\frac x2\right)^{2k}\left[1+2^k\frac1{(1-\tan^2(x/2))^k}\right]=\sec^{2n}\frac x2+\sec^n x.$$