蒜蓉包

題目係:In the expansion of (x-1/x)^10*(x+1/4)^4, find
(a) the constant term
(b) the coefficient of x^12

我知道答案係 (a) -72 (b) -6, 但唔明點解我用個個方法有問題


(a) For (x-1/x)^10, the general term = 10Cr * x^(10-r) * (-1/x)^r = 10Cr * x^(10-2r) * (-1)^r
For (x+1/4)^4, the general term= 4Cr * x^(4-2r)

跟住我就諗，如果係constant term，咁兩個x個power乘埋應該等如0，咁即係
10-2r + 4 - 2r = 0
r = 7/2


但7/2唔係integer，咁即係冇term係啱，於是試埋(b)

(b) When 10-2r+4-2r=12
r= 1/2
又唔得

我估計應該係我concept有錯，雖然睇得明model answer，但想知道除左個個方法以外仲有冇其他，general term係咪唔work？定要睇d咩情況？

jwyl

general term applies only on ONE expansion such as  (x-1/x)^10 or (x+1/4)^4 but for (x-1/x)^10*(x+1/4)^4, use the method in model answer.

AZA

(a) For (x-1/x)^10, the general term = 10Cr * x^(10-r) * (-1/x)^r = 10Cr * x^(10-2r) * (-1)^r
For (x+1/4)^4, the general term= 4Cr * x^(4-2r)

10-2r + 4 - 2r = 0 <-實際上要分r1,r2=2個unknown
因為兩舊個r可以唔同
所以最簡單都係拆開佢

emmanet10

樓主計法, 假設 (x-1/x)^10 相乘 (x+1/4)^4完,  係無其他constant term 出現
事實上, 唔係!

[ 本帖最後由 emmanet10 於 2014-1-12 09:28 PM 編輯 ]

蒜蓉包

咁即係面對呢d題型就要無奈得拆？

蒜蓉包

如果依照你所講r1 r2咁分兩個unknown就計無可計?

AZA

兩舊就用唔到general form 就咁計r
不排除有其他好D嘅方法(有就等其他高手教路啦 )
不過最直接都係拆開哂佢 至少比你慢慢搵其他方法快

蒜蓉包

原來係咁
咁個model answer做既就只係簡化+拆，萬一睇唔到個個(a+b)(a-b)既話，就咁拆係米都會計到？

AZA

就咁拆一定計到 不過就煩少少

cwk7093942

form your idea

cwk7093942

part b) , from the above

you can use the same idea to find the solution sets of (k ,  r) = (9 , 4) and (10 , 3)

you will easily to get the coefficient of x^12 = -6

蒜蓉包

oh, thank you so much. so from this idea, i can actually use two unknowns, and see which cases satifies the required sum, can this be applied to when there are more than two expressions?

cwk7093942

引用:

原帖由 蒜蓉包 於 2014-1-13 04:02 PM 發表
oh, thank you so much. so from this idea, i can actually use two unknowns, and see which cases satifies the required sum, can this be applied to when there are more than two expressions?



-mobi ...

方法就係上面...你可以自己試...

唔需要乜都靠人講....

依家唔係考試, 當你越嘗試得多, 對你越有好處...

蒜蓉包

係啦大師