用裂项法:
=[1/(2×4)+1/(4×6)+1/(6×8)+1/(8×10)+1/(10×12)+1/(12×14)+1/(14×16)+1/(16×18)]×128
=(1/2)(1/2-1/4+1/4-1/6+1/6-1/8+...+1/16-1/18)×128
=(1/2)×(1/2-1/18)×128
=(1/2)×(4/9)×128
=(2/9)×128
=256/9
与你前面问的几个题目是同一类型!
1/8=1/(2×4)=(1/2)×[(1/2)-(1/4)]
1/24=1/(4×6)=(1/2)×[(1/4)-(1/6)]
……
1/288=1/(16×18)=(1/2)×[(1/16)-(1/18)]
所以,括号内连续相加=(1/2)×[(1/2)-(1/4)+(1/4)-(1/6)+……+(1/16)-(1/18)]
=(1/2)×[(1/2)-(1/18)]
=2/9
所以,原式=(2/9)×128=256/9