1/1X(x+1) + 1/(x+1)(x+2)+1/(x+2)(x+3)……+1/(x+1998)(x+1999) 当
1个回答
裂项.1/x(x+1)=1/x-1/(x+1),1/(x+1)(x+2)=1/(x+1)-1/(x+2),以此类推……
所以最后原式=1/x-1/(x+1999),代入x=1得1999/2000
相关问题
计算:[1x(x+1)+1(x+1)(x+2)+1(x+2)(x+3)+…+1(x+1998)(x+1999)
计算:[1x(x+1)+1(x+1)(x+2)+1(x+2)(x+3)+…+1(x+1998)(x+1999)
计算:[1x(x+1)+1(x+1)(x+2)+1(x+2)(x+3)+…+1(x+1998)(x+1999)
计算:[1x(x+1)+1(x+1)(x+2)+1(x+2)(x+3)+…+1(x+1998)(x+1999)
1/(1x2)+1/(2x3)+1/(3x4)+.1/(1998x1999)
{X1+X2=X2+X3=X3+X4=...=X1997+X1998=X1998+X1999=1 X1+X2+...+X
解方程组:x1+x2=x2+x3=x3+x4=……=x1997+x1998=x1998+x1999=1x1+x2+……+
二元一次方程组难题{x1+x2=x2+x3=x3+x4=…=x1997+x1998=x1998+x1999=1{ x1+
求解方程组,怎样做{x1+x2=x2+x3=x3+x4=...=x1997+x1998=x1998+x1999=1,{x
(1999x)平方-1998*2000x-1=0怎么得(1999^2x+1)(x-1)=0