1. Find the coefficient of x^3 in the expansion of (10-7x)(1+x/5)^x
2. In the expansion of (1-2x)^n the sum of the coefficients of x and x^2 is 16 . Given that n is positive , find the value of n ,the the coefficient of x^3.
3.Given that the coefficient of x^2 in the expansion of (k+x)(2-x/2)^6 is 84 , find the value of the constant k
Coefficients
2010-04-21 3:35 am
回答 (4)
2010-04-21 6:46 pm
✔ 最佳答案
====================================================圖片參考：http://i826.photobucket.com/albums/zz190/fai109/421.jpg
2010-04-21 13:32:20 補充：
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2010-04-21 8:53 pm
Q1. If (1 + x/5)^x is correct, according to WolframAlpha, expansion is 1 + x^2/5 - x^3/50 + .......
so (10 - 7x)(1 + x/5)^x = (10 -7x)(1 + x^2/5 - x^3/50 + .....).
so coefficient of x^3 term = (10)(-1/50) + (-7)(1/5) = -1/5 - 7/5 = - 8/5.
Q2.
(1 - 2x)^n = 1 + n(-2x) + n(n -1)(-2x)^2/2! + .....
= 1 - 2nx + 2n(n-1)x^2 + .....
so - 2n + 2n(n-1) = 16
-n + n^2 - n - 8 = 0
n^2 - 2n - 8 = 0
(n - 4)(n + 2) = 0
n = - 2(rej.)
so n = 4.
The x^3 term is n(n-1)(n-2)(-2x)^3/3! = (4)(3)(2)(-8x^3)/6
so coefficient is - 32.
Q3.
(k + x)(2 - x/2)^6 = (k + x)[2^6 + 6(2^5)(-x/2) + 15(2^4)(-x/2)^2 + .....)
so coefficient of x^2 term is 15k(16)(1/4) + 6(32)(-1/2) = 84
60k - 96 = 84
60k = 180
k = 3.
so (10 - 7x)(1 + x/5)^x = (10 -7x)(1 + x^2/5 - x^3/50 + .....).
so coefficient of x^3 term = (10)(-1/50) + (-7)(1/5) = -1/5 - 7/5 = - 8/5.
Q2.
(1 - 2x)^n = 1 + n(-2x) + n(n -1)(-2x)^2/2! + .....
= 1 - 2nx + 2n(n-1)x^2 + .....
so - 2n + 2n(n-1) = 16
-n + n^2 - n - 8 = 0
n^2 - 2n - 8 = 0
(n - 4)(n + 2) = 0
n = - 2(rej.)
so n = 4.
The x^3 term is n(n-1)(n-2)(-2x)^3/3! = (4)(3)(2)(-8x^3)/6
so coefficient is - 32.
Q3.
(k + x)(2 - x/2)^6 = (k + x)[2^6 + 6(2^5)(-x/2) + 15(2^4)(-x/2)^2 + .....)
so coefficient of x^2 term is 15k(16)(1/4) + 6(32)(-1/2) = 84
60k - 96 = 84
60k = 180
k = 3.
2010-04-21 1:43 pm
STEVIE-G™，點解你覺得要展開(1 + x/5)^x就很有問題？
我都試過展開x^x（http://hk.knowledge.yahoo.com/question/question?qid=7008030602710），更何況在學e的時候都試過展開(1 + 1/x)^x，因此展開(1 + x/5)^x也是絕對可行的。
我都試過展開x^x（http://hk.knowledge.yahoo.com/question/question?qid=7008030602710），更何況在學e的時候都試過展開(1 + 1/x)^x，因此展開(1 + x/5)^x也是絕對可行的。
2010-04-21 5:59 am
(1+x/5)^x???
收錄日期: 2021-04-22 00:48:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100420000051KK01312