按x的升冪序，展開下列各式x^3項。
(1-4x+4x^2)^5
A-maths!!!!!!!!!
2007-06-19 6:17 am
回答 (4)
2007-06-19 6:28 am
✔ 最佳答案
(1 - 4x + 4x^2)^5= [ 1 + 4x(x - 1) ]^5
= 1 + (5C1)4x(x - 1) + (5C2)16x^2(x - 1)^2 + (5C3)64x^3(x - 1)^3 + ...
= 1 + 20x^2 - 20x + 160x^2(x^2 - 2x + 1) + 640x^3(x^3 + ... - 1) + ...
= 1 + 20x^2 - 20x - 320x^3 + 160x^2 - 640x^3 + ...
= 1 - 20x + 180x^2 - 960x^3 + ...
2007-06-19 9:18 am
豬油包's answer above is correct. However, it should be more nicely presented.
E.g.(1) Write everything in ascending power (don't write "(x - 1)" which is descending power!) as the question suggests the final answer to be.
E.g.(2) Add brackets to separate the many terms into groups.
E.g.(1) Write everything in ascending power (don't write "(x - 1)" which is descending power!) as the question suggests the final answer to be.
E.g.(2) Add brackets to separate the many terms into groups.
2007-06-19 7:21 am
(1 - 4x + 4x^2)^5
= [ 1 + 4x(x - 1) ]^5
= 1 + (5C1)4x(x - 1) + (5C2)16x^2(x - 1)^2 + (5C3)64x^3(x - 1)^3 + ...
= 1 + 20x^2 - 20x + 160x^2(x^2 - 2x + 1) + 640x^3(x^3 + ... - 1) + ...
= 1 + 20x^2 - 20x - 320x^3 + 160x^2 - 640x^3 + ...
= 1 - 20x + 180x^2 - 960x^3 + ...
= [ 1 + 4x(x - 1) ]^5
= 1 + (5C1)4x(x - 1) + (5C2)16x^2(x - 1)^2 + (5C3)64x^3(x - 1)^3 + ...
= 1 + 20x^2 - 20x + 160x^2(x^2 - 2x + 1) + 640x^3(x^3 + ... - 1) + ...
= 1 + 20x^2 - 20x - 320x^3 + 160x^2 - 640x^3 + ...
= 1 - 20x + 180x^2 - 960x^3 + ...
參考: from the first one
2007-06-19 6:31 am
(1-4x+4x^2)^5
=[1-x(4-4x)]^5
=(5C0)(1)^5-(5C1)[x(4-4x)]+(5C2)[x(4-4x)]^2-(5C3)[x(4-4x)]^3.................
=1-20x+20x^2+10x^2(16-32x+16x^2)-10x^3(64+....)+.......
=1-20x+20x^2+160x^2-32x^3-640x^3+....
=1-20x+180x^2-682x^3+........
=[1-x(4-4x)]^5
=(5C0)(1)^5-(5C1)[x(4-4x)]+(5C2)[x(4-4x)]^2-(5C3)[x(4-4x)]^3.................
=1-20x+20x^2+10x^2(16-32x+16x^2)-10x^3(64+....)+.......
=1-20x+20x^2+160x^2-32x^3-640x^3+....
=1-20x+180x^2-682x^3+........
收錄日期: 2021-04-23 20:36:17
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