Binomial Expansion(4)
2010-02-05 12:39 am
It is given that
(1+x+ax^2)^6 = 1+6x+k1x^2+k2x^3+terms involving higher powers of x.
(a) Express k1 and k2 in terms of a
(b) If 6, k1 and k2 are in arithmetic sequence, find the value of a.
回答 (2)
2010-02-05 1:03 am
✔ 最佳答案
a)(1+x+ax^2)^6
= (1 + x(ax + 1))^6
= 1 + [ 6C1 (x)(ax+1) + 6C2 (x^2)(ax+1)^2 + 6C3 (x^3)(ax+1)^3 + ... ]
= 1 + [ 6ax^2 + 6x + 15(x^2)(...+ 2ax + 1) + 20(x^3)(...+ 1) + ... ]
= 1 + [ 6ax^2 + 6x + 30ax^3 + 15x^2 + 20x^3 + ... ]
= 1 + 6x + (6a+15)x^2 + (30a+20)x^3 + ...
k1 = 3(2a + 5)
k2 = 10(3a + 2)
b) k1 - 6 = k2 - k1
6a+15-6 = 30a+20-(6a+15)
6a + 9 = 24a + 5
18a = 4
a = 2/9
2010-02-05 1:18 am
x^2係數
設 1取p次，x取q次，ax^2 取r次
=>p+q+r=6，q+2r=2
(1)r=0 =>q=2，p=4
係數和=6!/(4!2!0!)*(1)^4*(1)^2*(a)^0=15
(2)r=1 =>q=0，p=5
係數和=6!/(5!0!1!)*(1)^5*(1)^0*(a)^1=6a
k1=15+6a
2010-02-04 17:18:29 補充：
x^3係數
設 1取u次，x取v次，ax^2 取w次
=>u+v+w=6，v+2w=3
(1)w=0 =>v=3，u=3
係數和=6!/(3!3!0!)*(1)^3*(3)^3*(a)^0=20
(2)w=1 =>v=1，u=4
係數和=6!/(4!1!1!)*(1)^4*(1)^1*(a)^1=30a
k2=20+30a
設 1取p次，x取q次，ax^2 取r次
=>p+q+r=6，q+2r=2
(1)r=0 =>q=2，p=4
係數和=6!/(4!2!0!)*(1)^4*(1)^2*(a)^0=15
(2)r=1 =>q=0，p=5
係數和=6!/(5!0!1!)*(1)^5*(1)^0*(a)^1=6a
k1=15+6a
2010-02-04 17:18:29 補充：
x^3係數
設 1取u次，x取v次，ax^2 取w次
=>u+v+w=6，v+2w=3
(1)w=0 =>v=3，u=3
係數和=6!/(3!3!0!)*(1)^3*(3)^3*(a)^0=20
(2)w=1 =>v=1，u=4
係數和=6!/(4!1!1!)*(1)^4*(1)^1*(a)^1=30a
k2=20+30a
收錄日期: 2021-04-21 22:08:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100204000051KK00825