求(1-3x+4x^2)^5展開式中，x^3的係數 (10點)
回答 (2)
2007-09-18 4:29 am
✔ 最佳答案
(1 – 3x + 4x2)5 = [1 – x(3 – 4x)]5
= 1 + 5C1(1)[-x(3 – 4x)] + 5C2(1)[-x(3 – 4x)]2 + 5C3(1)[-x(3 – 4x)]3 + …
= 1 – 15x + 20x2 + 10x2(9 – 12x + …) - 270x3 + …
= 1 – 15x + 110x2 – 390x3 + …
所以，x3的係數是390
參考: Myself~~~
2007-09-18 4:31 am
(1 - 3x + 4x^2)^5
= [1 + (-3 + 4x)x]^5
= 1 + 5(-3 + 4x)x + 10(-3 + 4x)^2 x^2 + 10(-3 + 4x)^3 x^3 + ...
= ... + 10(-3 + 4x)^2 x^2 + 10(-3 + 4x)^3 x^3 + ...
= ... + 10(-24x + ..) x^2 + 10(-27 + ...) x^3 + ...
= ... + (-240 - 270) x^3 + ...
= ... + (-510) x^3 + ...
x^3 的係數 = -510
2007-09-17 20:33:48 補充：
上面位人兄好似做錯少少野...
= [1 + (-3 + 4x)x]^5
= 1 + 5(-3 + 4x)x + 10(-3 + 4x)^2 x^2 + 10(-3 + 4x)^3 x^3 + ...
= ... + 10(-3 + 4x)^2 x^2 + 10(-3 + 4x)^3 x^3 + ...
= ... + 10(-24x + ..) x^2 + 10(-27 + ...) x^3 + ...
= ... + (-240 - 270) x^3 + ...
= ... + (-510) x^3 + ...
x^3 的係數 = -510
2007-09-17 20:33:48 補充：
上面位人兄好似做錯少少野...
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