3/1*2+3/2*3+3/3*4+…+3/2003*2004计算:3/1*2+3/2*3+3/3*4+…+3/2003*2004.

问题描述:

3/1*2+3/2*3+3/3*4+…+3/2003*2004
计算:3/1*2+3/2*3+3/3*4+…+3/2003*2004.

通项公式:
1/[n(n+1)]=1/n-1/(n+1)
3/1*2+3/2*3+3/3*4+…+3/2003*2004
=3*[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2003-1/2004)]
=3*(1-1/2004)
=3*2003/2004

原式=3(1-1/2+1/2-1/3+……+1/2002-1/2004)
=3(1-1/2004)
=3*2003/2004
=2399/800

用通项公式:
1/[n(n+1)]=1/n-1/(n+1)
既有
3/1*2+3/2*3+3/3*4+…+3/2003*2004
=3*[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2003-1/2004)]
=3*(1-1/2004)
=3*2003/2004 =3[1-1/2400]=2399/800

3[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)++…+(1/2003-1/2004)]=3[1-1/2400]=2399/800

1/1*2=1/1-1/2
1/2*3=1/2-1/3
1/3*4=1/3-1/4
.
1/2003*2004=1/2003 - 1/2004
∴原式=3[1/1*2 +1/2*3 +1/3*4 +...+1/2003*2004]
=3[1-1/2 +1/2-1/3 +1/3-1/4 +...+1/2003 -1/2004]
=3[1-1/2004]=3*2003/2004=6009/2004

3(1-1/2+1/2-1/3+……+1/2002-1/2004)
=3(1-1/2004)
=3*2003/2004
=2又667/668