1/3×7 =(1/3-1/7)/4
1/7×11 =(1/7-1/11)/4
....
1/55×59=(1/55-1/59)/4
相加过程中,前后2项可以抵消,结果:
1/3×7 +1/7×11 +1/11×15 +······1/55×59
=(1/3-1/59)/4
对于这种分母为有规律的 m(m+b), (m+b)(m+2b)其中b常数,通过抵消前后2项的(m+b),等式最后等于(1/第一项-1/最后一项)/b,都可以通过这样来计算
化成两个分式相减~~
列项求和
1/3×7 +1/7×11 +1/11×15 +······1/55×59
=(1/4)[1/3-1/7+1/7-1/11+.……+1/51-1/55+1/55-1/59]
=(1/4)[1/3-1/59]
=(1/4)[18/59]
=9/118
1/3×7 +1/7×11 +1/11×15 +······1/55×59
=1/4×(1/3-1/7)+1/4×(1/7-1/11)+1/4×(1/11-1/15)+......+1/4×(1/55-1/59)
=1/4×(1/3-1/7+1/7-1/11+1/11-1/15+......+1/55-1/59)
=1/4×(1/3-1/59)
=1/4×56/177
=14/177