Binomial

你識既a.a (咪扮唔識啦!)

攪笑......

就是力量。

2010-04-27 15:01:58 補充：
a) (1 + ax + bx^2)^8

= [1 + (ax + bx^2)]^8

= 1 + 8C1 (ax + bx^2) + 8C2 (ax + bx^2)^2 + 8C3 (ax + bx^2)^3 + ....

= 1 + 8(ax + bx^2) + 28(a^2 x^2 + 2abx^3 + ....) + 56(a^3 x^3 + ....) + ....

= 1 + 8ax + (28a^2 + 8b)x^2 + (56a^3 + 56ab)x^3 + ....



b) Using the result obtained in (a)

(1 + mx + nx^2)^8 = 1 + 8mx + (28m^2 + 8n)x^2 + (56m^3 + 56mn)x^3 + ....

for

(1 + mx + nx^2)^8 = 1 + 4x + 0x^2 + (terms involving x^3 and higher powers of x) ......☆

Comparing coefficients , 8m = 4 ,

m = 1/2

and 28m^2 + 8n = 0

Substituting m = 1/2

28(1/2)^2 + 8n = 0

n = - 7/8


c) Putting x = 0.016 and neglecting terms involving x^3 and higher powers of x to ☆
[1 + m(0.016) + n(0.016)^2] ^ 8 = 1 + 4(0.016)

Substituting the values of m and n,

[1 + (1/2)(0.016) + (-7/8)(0.016)^2] ^ 8 = 1.064

1 + 0.008 - 0.000224 = 1.064 ^ (1/8)

1.064 ^ (1/8) = 1.007776 = 1.00778(correct to 5 places of decimals)

你識既a.a

2010-04-26 20:56:15 補充：
傻了-_-
打漏了很多東西,
(b) ....... higher powers of x. Find the values of m and n

(1+ax+bx^2)^8
= [1+x(a+bx)]^8
= 1 + 8x(a+bx) + 28x^2 (a+bx)^2 + 56 x^3 (a+bx)^3 + ...
= 1 + 8ax + 8bx^2 + 28a^2 x^2 + 56abx^3 + 56a^3 x^3 + ...
= 1 + 8ax + (8b+28a^2)x^2 + (56ab+56a^3)x^3 + ...
其他唔識 = =|||
小小心意 ^^"