Tính tổng \(S=C_{2n}^0+C^2_{2n}+.......+C^{2n}_{2n}\) .
\(2^{n-1}\).\(2^n\).\(2^{n+1}\).\(2^{n-2}\).Hướng dẫn giải:Xét khai triển \(\left(x+1\right)^{2n}=C^0_{2n}x^0+C^1_{2n}x+C^2_{2n}x^2+..........+C^{2n-1}_{2n}x^{2n-1}+C^{2n}_{2n}x^{2n}\).
Thay \(x=1\), ta có: \(2^{2n}=C^0_{2n}+C^1_{2n}+C^2_{2n}+..........+C^{2n-1}_{2n}+C^{2n}_{2n}\).
Thay \(x=-1\), ta có: \(0=C^0_{2n}-C^1_{2n}+C^2_{2n}+..........-C^{2n-1}_{2n}+C^{2n}_{2n}\).
Suy ra \(2^{2n}=2\left(C^0_{2n}+C^2_{2n}+......+C^{2n}_{2n}\right)\) hay \(C^0_{2n}+C^2_{2n}+C^4_{2n}+....+C^{2n}_{2n}=2^{n-1}\).