(X+2)^2(X-3)^8 。 求第六項的係數 和常項數
關於"二項展式"的問題
2011-01-28 6:15 am
回答 (2)
2011-01-28 6:40 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_02Jan272239.gif
2011-01-28 6:30 am
(x+2)^2 (x-3)^8
= (x^2+4x+4)(3 - x)^8
= (x^2+4x+4)[3^8 - 8(3^7)(x) + 28(3^6)(x^2) - 70(3^5)(x^3) + 56(3^4)(x^4) - 28(3^3)(x^5) + .....]
= (x^2+4x+4)(6561 - 17496x + 20412x^2 - 17010x^3 + 4536x^4 - 756x^5 + ...)
第六項的係數
= (4536)(1) + (-756)(4)
= 1512
常項數
= 4(6561)
= 26244
= (x^2+4x+4)(3 - x)^8
= (x^2+4x+4)[3^8 - 8(3^7)(x) + 28(3^6)(x^2) - 70(3^5)(x^3) + 56(3^4)(x^4) - 28(3^3)(x^5) + .....]
= (x^2+4x+4)(6561 - 17496x + 20412x^2 - 17010x^3 + 4536x^4 - 756x^5 + ...)
第六項的係數
= (4536)(1) + (-756)(4)
= 1512
常項數
= 4(6561)
= 26244
參考: Knowledge is power.
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