(1/2+1/3+…1/2003)(1+1/2+…+1/2002)-(1+1/2+…+1/2003)(1/2+1/3+…
1个回答
记t=1+1/2+…+1/2003
原式=(t-1)(t-1/2003)-t(t-1-1/2003)
=t^2-t-t/2003+1/2003-t^2+t+t/2003
=1/2003
相关问题
简便计算:1/2+1/3+2/3+1/4+2/4+3/4+.+1/2003+2/2003+.+2002/2003
1/(21*2)+1/(2*3)···+1/(2002*2003)+1/(2003*2004)+1/(2004*2005
计算:1+2+3+…+2002+2003+2002+…+3+2+1=______.
1+2+3+4+...+2002+2003+2002+2002+...+3+2+1=?
1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+4+5+.+2002+2003)=?=
1+1/(1+2)+1/(1+2+3)+.+1/(1+2+3+4+5+.+2002+2003)=?=
1.2005*2004-2004*2003+2003*2002-2002*2001+..+3*2-2*1
找规律1+2+3+...2002+2003+2002+...+3+2+1=
(1-1/2004)(1-2003)(1-2002)…(1-1/3)(1-1/2)
(1/2+1/3...+1/2003)(1+1/2+1/3+.+1/2004)-(1+1/2+1/3+.+1/2003)