利用1/n*(n+1) = 1/n - 1/(n+

利用1/n*(n+1) = 1/n - 1/(n+1)计算1/（x-2)（x-3)-2/(x-1)(x-3)+1/(x-

2个回答

1/（x-2)（x-3)-2/(x-1)(x-3)+1/(x-1)(x-2)
=1/(x-3)-1/(x-2)-[1/(x-3)-1/(x-1)]+1/(x-2)-1/(x-1)
=1/(x-3)-1/(x-2)-1/(x-3)+1/(x-1)+1/(x-2)-1/(x-1)
=0

计算(x^(2n)+x^n+1)(x^(3n)-x^(2n)+1)
Sn=1/(1x2x3)+1/(2x3x4)+...+1/(n(n+1)(n+2))
x^(3n+1)y^(n-1)+2x^(2n+1)y^(2n-1)+x^(n+1)y^(3n-1)因式分解
（1-x)Sn=1+2x[1+x+x^2+x^3+…+x^(n-2)]-(2n-1)*x^n；
\求和Sn=1+2x+3x^2+```+(n-1)x^(n-2)+n*x^(n-1)
1/1x2x3+1/2x3x4十……十1/n(n+1)(n十2)
计算：-4(1/2x-2)+1/3(6x-9)；3n-[5n+(3n-1)]
计算:(8x^n+3 + x^n+2) * (-2x^n + 3x^n-1)
(x+1)^n=a0+a1(x-1)+a2(x-1)^2+a3(x-1)^3+.+an(x-1)^n ,(n>=2n属于
1X2X3+2X3X4+.+n(n+1)(n+2)