a-maths~~~~~~~~

(1 – 2x)n

= 1 + nC1(1)(-2x) + nC2(1)(-2x)2 + nC3(1)(-2x)3 + nC4(1)(-2x)4 + …

= 1 – 2nx + 2n(n - 1)x2 - 4n(n - 1)(n - 2) / 3 x3 + 2n(n - 1)(n - 2)(n - 3) / 3 x4 + …

x4的係數 : x2的係數 = 20 : 3

[2n(n - 1)(n - 2)(n - 3) / 3] : 2n(n - 1) = 20 : 3

(n - 2)(n - 3) = 20

n2 – 5n - 14 = 0

(n - 7)(n + 2) = 0

n = 7 或 -2 (捨去)