北京昌平etc客服电话:一道数学题,拜托大家帮一下忙吧

预备知识:
以下这个数字阵横行观察,行数表达的是n的取值,数字表示的是(a+b)^n展开式中按a的降幂排列后各个项的系数.

..............1...1..............n=1
............1...2...1............n=2
..........1...3...3...1..........n=3
........1...4...6...4...1........n=4
......1...5..10..10...5...1......n=5
.....1...6..15..15..15...6...1....n=6
...................................n

所以,
(a+b)^2=a^2+2ab+b^2
(a+b)^3=a^3+3a^2b+3ab^2+b^3
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5
即有:
a^2+b^2=(a+b)^2-2ab
a^3+b^3=(a+b)^3-3a^2b-3ab^2
a^4+b^4=(a+b)^4-4a^3b-6a^2b^2-4ab^3
a^5+b^5=(a+b)^5-5a^4b-10a^3b^2-10a^2b^3-5ab^4

其他理论根据:
sin^2(θ) + cos^2(θ) =1
2sinθcosθ=sin2θ
为了书写方便,
设
sin^2(θ)=A
cos^2(θ)=B
sin^2(2θ)=C
则有
A+B=1
AB=(1/4)*C
A^2+B^2=(A+B)^2-2AB=1-(1/2)*C
u6
=A^3+B^3
=(A+B)^3-3A^2B-3AB^2
1-3AB(A+B)
=1-(3/4)*C
u8
=A^4+B^4
=(A+B)^4-4A^3B-6A^2B^2-4AB^3
=1-4AB(A^2+B^2)-6(AB)^2
=1-4AB[1-(1/2)*C]-6(AB)^2
=1-4*(1/4)*C*[1-(1/2)*C]-(6/16)*C^2
=1-C+(1/2)C^2-(3/8)*C^2
=1-C+(1/8)*C^2
u10
=A^5+B^5
=(A+B)^5-5A^4B-10A^3B^2-10A^2B^3-5AB^4
=1-5AB(A^3+B^3)-10A^2B^2(A+B)
=1-5AB[1-(3/4)*C]-10*(1/16)*C^2
=1-5*(1/4)*C[1-(3/4)*C]-(5/8)*C^2
=1-(5/4)*C*[1-(3/4)*C]-(5/8)*C^2
=1-(5/4)*C+(15/16)*C^2-(5/8)*C^2
=1-(5/4)*C+(5/16)*C^2

即:
u6 =1-(3/4)*C
u8 =1-C+(1/8)*C^2
u10=1-(5/4)*C+(5/16)*C^2

6u10-15u8+10u6
=6*[1-(5/4)*C+(5/16)*C^2]-15*[1-C+(1/8)*C^2]+10*[1-(3/4)*C]
=6-(15/2)*C+(15/8)*C^2-15+15C-(15/8)*C^2+10-(15/2)*C
=6-15+10
=1
所以6u10-15u8+10u6的结果与θ的取值无关.

希望能帮到你. :)

由Sin2(a)+Cos2(a)=1 （希望看得懂，这样写我感觉比较方便）
设Sin2(a)=x, Cos2(a)=y
原式= 6*x^5 +6*(1-x)^5 -15*x^4 -15*(1-x)^4 +10*x^3 +10*(1-x)^3
下面将(1-x)^i展开，用贾宪三角形，别说不会
原式=6*x^5-6(x^5-5*x^4+10*x^3-10*x^2+5*x-1)-15*x^4-15(x^4-4*x^3+6*x^2-4*x+1)+10*x^3-10(x^3-3*x^2+3*x-1)
=6*x^5-6*x^5+30*x^4-60*x^3+60*x^2-30*x+6-15*x^4-15*x^4+60*x^3-90*x^2+60*x-15+10*x^3-10*x^3+30*x^2-30*x+10
=1 （常数）

好难......慢慢推吧