It is given that
(1 - 2x + 3x^2)^n = 1 - 10x + kx^2 = + terms involving higher powers of x, where n is a positive interger and k is a constant. Find the values of n and k.
A maths 問題
2008-03-29 3:24 am
回答 (1)
2008-03-29 3:52 am
✔ 最佳答案
(1-2x+3x^2)^n=1-10x+kx^2 + terms involving higher powers of x,(1+x(3x-2))^n=1-10x+kx^2 + terms involving higher powers of x
1+nx(3x-2)+[(n(n-1)/2][x(3x-2)]^2+...=1-10x+kx^2 + terms involving higher powers of x
1+nx(3x-2)+[(n(n-1)/2][x(3x-2)]^2=1-10x+kx^2+...
1+3nx^2-2nx+[(n(n-1)/2][4x^2]=1-10x+kx^2
1+3nx^2-2nx+[(n(n-1)][2x^2]=1-10x+kx^2
-2n=-10
3n+2(n)(n-1)=k
n=5
k=15+2*5*4
k=55
收錄日期: 2021-04-26 13:04:34
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