1乘2分之1加2乘3分之一一直加到2013乘2014分之一
1/1x2 + 1/2x3 + 1/3x4 + ... + 1/2013x2014
= (1/1 - 1/2) + (1/2 - 1/3) + (1/3 -1/4) + ... + (1/2012 - 1/2013) + (1/2013 - 1/2014)
= 1/1 -1/2+1/2 -1/3+1/3 - 1/4+1/4 ... - 1/2013+1/2013 - 1/2014
= 1/1 - 1/2014
= 2013/2014
此类题目的一般解题思路就是裂项:
裂项的基本原理是,假设正整数b>a,则1/a x b =(1/a - 1/b) * (b-a)
当b -a = 1时,裂项公式特例化为1/a x b = 1/a - 1/b
常用裂项相消法公式:
(1)1/[n(n+1)]=(1/n)- [1/(n+1)]
(2)1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
(3)1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
(4)1/(√a+√b)=[1/(a-b)](√a-√b)
(5) n·n!=(n+1)!-n!
(6)1/[n(n+k)]=1/k[1/n-1/(n+k)]
(7)1/(√n+√n+1)=√(n+1)-√n
(8)1/(√n+√n+k)=(1/k)·[√(n+k)-√n]