Course Content
高二数学 | 高级数学
About Lesson
7.1 Binomial Theorem with Natural Number Power 指数为自然数的二项式定理


[1]

Expand the following binomials :
展开以下的二项式

(a) (x – 3)^5

(b) (3x + 2y)^4

(c) (1 – 3x^2)^4

(d) (x – 2)(3x + 1)^3

(e) \left(\sqrt x-\frac{1}{2\sqrt x}\right)^5

Answer:
(a) x^5 - 15x^4 + 90x^3270x^2 + 405x - 243

(b) 81x^4 + 216x^3y + 216x^2y^2 + 96xy^3 + 16y^4

(c) 1 - 12x^2 + 54x^4108x^6 + 81x^8

(d) 27x^4 - 27x^3 - 45x^217x - 2

(e) x^\frac{5}{2}-\frac{5}{2}x^\frac{3}{2}+\frac{5}{2}x^\frac{1}{2}\frac{5}{4}x^{-\frac{1}{2}}+\frac{5}{16}x^{-\frac{3}{2}}-\frac{1}{32}x^{-\frac{5}{2}}

[2]

Find in ascending powers of t, the first three terms in the expansions of
以t的升幂方式展开以下各式的首三项

(a) (1 + {\alpha}t)^5

(b) (1 - {\beta}t)^8

Hence find, in terms of \alpha, and \beta, the coefficient of t^2 in the expansion of (1 + \alphat)^5(1 – \betat)^8.

以此求, 以\alpha\beta表示, 在(1 + \alphat)^5(1 – \betat)^8的展开式中t^2的系数.

Answer:
(a) 1 + 5\alphat + 10\alpha2t^2 + …

(b) 1 – 8\betat + 28\beta2t^2 – …

10\alpha^2 + 28\beta^2 – 40\alpha\beta

[3]

If (1 + 2x + 3x^2)^5 = a_0+ a_{1}x + a_{2}x^2 + … + a_{10}x^{10}. calculate

若 (1 + 2x + 3x^2)^5 = a_0+ a_{1}x + a_{2}x^2 + … + a_{10}x^{10}

(a) a_0+ a_1 + a_2 + … + a_{10}

(b) a_0+ a_2 + a_4 + … + a_{10}

Answer:
(a) 7776

(b) 3904