F.4 AMaths----Binomial Theorm

1) (1-x)^7 + x(1-x)^8 + x^2(1-x)^9

the coefficient of x^5 = (-1)^5 (7C5) + (-1)^4 (8C4) + (-1)^3 (9C3)
= -21+70-84
= -35
要求到x^5,就要睇下邊度乘出泥係會有x^5
(1-x)^7 好明顯就係第6個term,唔好忘記總共有8個term (由7C0去到7C7總共有8個數)
咁第6個term就係 (-1)^5 (7C5)
到x(1-x)^8,因為每個term都要乘返x,所以只需要搵返個x^4 就得啦
咁第5個term就係 (-1)^4 (8C4)
到x^2(1-x)^9,今次要乘x^2,所以搵x^3 就得啦
咁第4個term就係 (-1)^3 (9C3)
咁就計到啦

2) 呢題好似難d咁,但只要記得個原理就都係一樣咁做架姐
先搵第7個term
即係話 (-1)^6 (nC6) (2x^3)^(n-6) [1/(3x)]^6
咁呢個係constant term, 即係話唔會有x係度
咁即係前後2個x係會互相抵消
所以 3(n-6)=6 (前後x會抵消就要出現咁既結果)
n=8
咁將n=8放返入條式度
constant term = (-1)^6 8C6 2^2 (1/3)^6
=112/729

希望你明白
邊度有唔明可以再問我既

1)Find the coefficient of x5 in the expansion of
(1-x)7 + x(1-x)8 + x2(1-x)9
(1-x)7中的x5係數
-7C5 = -21
x(1-x)8中的x5係數
8C4 = 70
x2(1-x)9中的x5係數
-9C3 = -84
所以(1-x)7 + x(1-x)8 + x2(1-x)9 的x5 係數
-21 + 70 – 84
= -35

2)When (2x3 - 1/(3x))n is expanded in descending powers of x , the 7th term is a constant term.Find the values of n and the constant term.
(2x3 - 1/(3x))n
= a1x3n + a2x3n-4 + a3x3n-8 + a4x3n-12 + a5x3n-16 + a6x3n-20 + a7x3n-24 + ….
第七項為常數
即 3n – 24 = 0
n = 8
a7 = 8C6(2)2(-1/3)6
= 28*22*(1/36)
= 112 / 729