Binomial Expansion

知識就是力量

2010-02-04 12:59 am

(a) Expand (1-2x+3x^2)^7 in ascending powers of x as far as the term containing x^3.
(b) Hence, evaluate 0.9803^7 and correct your answer in 4 decimal places.

回答 (1)

用戶2053

2010-02-04 2:37 am

✔ 最佳答案

a)(1-2x+3x^2)^7
= (1 + (3x^2 - 2x) )^7
= 1 + 7C1 (3x^2 - 2x) + 7C2 (3x^2 - 2x)^2 + 7C3 (3x^2 - 2x)^3+..+(3x^2 - 2x)^7
= 1 + 7x(3x - 2) + (21x^2) (3x - 2)^2 + (35x^3) (3x - 2)^3 +...+(3x^2 - 2x)^7
= 1 + 21x^2 - 14x + (21x^2)(... - 12x + 4) + (35x^3)(.... - 2^3)+.+(3x^2-2x)^7
= 1 + 21x^2 - 14x - 252x^3 + 84x^2 - 280x^3 + ... + (3x^2 - 2x)^7
= 1 - 14x + 105x^2 - 532x^3 + ...
b) Let x be 0.01 , then 1 - 2x + 3x^2 = 1 - 0.02 + 0.0003 = 0.9803
0.9803^7 = 1 - 14(0.01) + 105(0.01)^2 - 532(0.01)^3 + ...
= 0.869968 = 0.8700 (4 dec.)

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