Biết rằng (3x – 1)^7 = a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + a7x7. Hãy tính: a) a0 + a1 + a2 + a3 + a4 + a5 + a6 + a7; b) a0 + a2 + a4 + a6.

Hướng dẫn giải

Có (3x – 1)⁷

$=C_7^0{\left(3x\right)}^7+C_7^1{\left(3x\right)}^6(-1)+C_7^2{\left(3x\right)}^5{(-1)}^2+C_7^3{\left(3x\right)}^4{(-1)}^3$

$+C_7^4{\left(3x\right)}^3{(-1)}^4+C_7^5{\left(3x\right)}^2{(-1)}^5+C_7^6{\left(3x\right)}^1{(-1)}^6+C_7^7{(-1)}^7$

= 2187x7 – 5103x6 + 5103x5 – 2835x4 + 945x3 – 189x2 + 21x – 1.

a) a₀ + a₁ + a₂ + a₃ + a₄ + a₅ + a₆ + a₇

= (–1) + 21 + (–189) + 945 + (–2835) + 5103 + (–5103) + 2187 = 128.

b) a₀ + a₂ + a₄ + a₆

= (–1) + (–189) + (–2835) + (–5103) = –8128.