二分之一加六分之一加十二分之一加二十分之一加三十分之一+42分之一+56分之一
=1/2+1/6+1/12+1/20+1/30+1/42+1/56
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=1-1/8
=7/8
=1/1×2+1/2×3+1/3×4+1/4×5+1/5×6+1/6×7+1/7×8
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
那么除了开头的“1”和结尾的“1/8”其他的都可以抵消,然后,就等于
=1-1/8
=7/8
裂项求和与倒序相加、错位相减、分组求和等方法一样,是解决一些特殊数列的求和问题的常用方法.这些独具特点的方法,就单个而言,确实精巧,
例子:
求和:
1/2+1/6+1/12+1/20+1/30+1/42+1/56
=1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5) +1/(5*6)+1/(6*7)+1/(7*8)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5) +(1/5-1/6)+(1/6-1/7)+(1/7-1/8)
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5 +1/5-1/6+1/6-1/7+1/7-1/8
=1-1/8=7/8
在裂项求和中最常见的是已知an(数列)求和。一般在高二数学中存有,是一类规律性题目。
1/2 1/6 1/12 1/20 1/30 1/42 1/56 =1/(1*2) 1/(2*3) 1/(3*4) 1/(4*5) 1/(5*6) 1/(6*7) 1/(7*8) =(1-1/2) (1/2-1/3) (1/3-1/4) (1/4-1/5) (1/5-1/6) (1/6-1/7) (1/7-1/8) =1-1/2 1/2-1/3 1/3-1/4 1/4-1/5 1/5-1/6 1/6-1/7 1/7-1/8 =1-1/8=7/8