Find constant term and coefficient of x^2y^3 in the expansion of (x+2/y)^4(y^2-5/x)^3 expand((x-5)(x-4))^4 in descending powers of x up to x^6 term
given that 1/8[(x+2/x^2)^6-(x-2/x^2)^6]=ax^3+b/x^3+c/x^9
a. find values of a,b,c
b. simplify 1/8[(root2+1)^6-(root2-1)^6]
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F.4 M2 Binomial Theorem
2012-09-11 5:04 am
回答 (1)
2012-09-11 5:06 pm
✔ 最佳答案
(x+2/y)^4 (y^2-5/x)^3= (x^4+8x^3/y+24x^2/y^2+32x/y^3+16/y^4)(y^6-15y^4/x+75y^2/x^2-125/x^3)
constant terms = 24x75 = 1800
coefficient of x^2y^3 = 8x(-15) = -120
((x-5)(x-4))^4
= (x-5)^4 (x-4)^4
= (x^4-20x^3+150x^2-500x+625)(x^4-16x^3+96x^2-256x+256)
= x^8-36x^7+566x^6+...
1/8[(x+2/x^2)^6-(x-2/x^2)^6]
= 1/4[6x^5(2/x^2)+20x^3(2/x^2)^3+6x(2/x^2)^5]
= 3x^3+40/x^3+48/x^9
1/8[(root2+1)^6-(root2-1)^6]
= 3(root2)^3+40/(root2)^3+48/(root2)^9
= 6root2+10root2+3/2root2
= 35/2root2
參考: knowledge
收錄日期: 2021-04-28 14:21:36
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