Sử dụng công thức nhị thức Newton, chứng tỏ rằng: a) C 0 4 + 2 C 1 4

a) \(C_4^0 + 2C_4^1 + {2^2}C_4^2 + {2^3}C_4^3 + {2^4}C_4^4 = 81\)

Ta có: \({\left( {1 + 2} \right)^4} = C_4^0{.1^4} + C_4^1{.1^3}.2 + C_4^2{.1^2}{.2^2} + C_4^3{.1^3}{.2^3} + C_4^4{.1.2^4}\)

⇔ \({3^4} = C_4^0 + C_4^1.2 + C_4^2{.2^2} + C_4^3{.2^3} + C_4^4{.2^4}\)

⇔ \(C_4^0 + 2C_4^1 + {2^2}C_4^2 + {2^3}C_4^3 + {2^4}C_4^4 = 81\) (đpcm).

b) \(C_4^0 - 2C_4^1 + {2^2}C_4^2 - {2^3}C_4^3 + {2^4}C_4^4 = 1\)

Ta có: (1 – 2)⁴ = \(C_4^0{.1^4} + C_4^1{.1^3}.( - 2) + C_4^2{.1^2}.{( - 2)^2} + C_4^3.1.{\left( { - 2} \right)^3} + C_4^4.{( - 2)^4}\)

⇔ (-1)⁴ = \(C_4^0 - C_4^1 + C_4^2{.2^2} - C_4^3{.2^3} + C_4^4{.2^4}\)

⇔ 1 = \(C_4^0 - 2C_4^1 + {2^2}C_4^2 - {2^3}C_4^3 + {2^4}C_4^4\) (đpcm).