1/x(x+3) + 1/(x+3)(x+6)+...+1/(x+2010)(x+2013)
=1/3 * [1/x -1/(x+3)]+1/3[1/(x+3) - 1/(x+6)]+...+1/3*[1/(x+2010) - 1/(x+2013)]
=1/3*[1/x - 1/(x+2013)]
=1/3* 2013/x(x+2013)
=2013/3x(x+2013)
1/x(x+3)+1/(x+3)(x+6)+...+1/(x+2010)(x+2013)
2个回答
相关问题
-
1/1x3+1/3x5+1/5x7.1/2009x2010+1/2011x2013
-
若X²+X-1=5求3X³+6X²-2013
-
2010x/1+2 + 2010x/ 1+2+3 + 2010x/1+2+3+4 + 2010x/1+2+.2010=2
-
x/1x2+x/2x3+x/3x4+.x/2010x2011=2010
-
化简1+x+x（x+1）+x（x+1）^2+x（x+1）^3+…+x（x+1）^2013
-
X1·X2·X3···X2010=1,且X1,X2,···X2010都是正数.（1+X1）（1+X2）····（1+X2
-
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2012)(x+2013)=
-
x1x2x3.x2011=1x1-x2x3.x2011=1x1x2-x3x4.x2011=1.x1x2.x2010-x2
-
x/1x2+x/2x3+.+x/2010x2011=2010
-
1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18