Question

Записать разложение бинома (2а-1/5)^5

Answer 1

(2a-1/5)^5=2a^5+5*2a^4*(-1/5)+10*2a^3*(-1/5)^2+10*2a^2*(-1/5)^3+5*2a*(-1/5)^4+(-1/5)^5=32a^5-16a^4+16/5a^3-8a^2/25+2a/125-1/3125.

Answer 2

(2a - 1/5)^5 =

C(5, 0)(2a)^5 * (-1/5)^0 + C(5, 1)(2a)^4 * (-1/5)^1 + C(5, 2)(2a)^3 * (-1/5)^2 + C(5, 3)(2a)^2 * (-1/5)^3 + C(5, 4)(2a)^1 * (-1/5)^4 + C(5, 5)(2a)^0 * (-1/5)^5

= 1 * (32a^5) * 1 + 5 * (16a^4) * (-1/5) + 10 * (8a^3) * (1/25) + 10 * (4a^2) * (-1/125) + 5 * (2a) * (1/625) + 1 * 1 * (-1/3125)

= 32a^5 - 16a^4 + 64a^3 - 8a^2 + 2a - 1/3125