裂项法是怎样的,例:1/3+1/15+1/35+1/63+1/99 该怎样算?

问题描述:

裂项法是怎样的,例:1/3+1/15+1/35+1/63+1/99 该怎样算?

1/3+1/15+1/35+1/63+1/99
=1/(4-1)+1/(16-1)+1/(36-1)+1/(64-1)+1/(100-1)
=[1/(2-1)-1/(2+1)]/2+[1/(4-1)-1/(4+1)]/2+[1/(6-1)-1/(6+1)]/2+[1/(8-1)-1/(8+1)]/2+[1/(10-1)-1/(10+1)]/2
=[1/1-1/3]/2+[1/3-1/5]/2+[1/5-1/7]/2+[1/7-1/9]/2+[1/9-1/11]/2
=[1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11]/2
=(1-1/11)/2
=5/11

楼主你好~
=1*1/3+1/3*1/5+1/5*1/7+1/7*1/9+1/9*1/11
=1/2[(1-1/3)+.......+(1/9-1/11)]
=1/2*(1-1/11)
=5/11

原式=1/2×(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=1/2×(1-1/11)
=5/11

裂项法:就是把一项拆成两项的差,前后抵消.
1/3+1/15+1/35+1/63+1/99
=(1/2)(1/1 -1/3)+(1/2)(1/3 -1/5)+(1/2)(1/5-1/7)+(1/2)(1/7-1/9)+(1/2)(1/9 -1/11)
=(1/2)(1-1/11)=5/11