Question

数列{a n }满足a 1 =1，且对任意的正整数m，n都有a m+n =a m +a n +mn，则 1 a 1 + 1

数列{a n }满足a 1 =1，且对任意的正整数m，n都有a m+n =a m +a n +mn，则 1 a 1 + 1 a 2 +…+ 1 a 2012 + 1 a 2013 =______．

Answer 1

令n=1，得a n+1 =a 1 +a n +n=1+a n +n，∴a n+1 -a n =n+1 用叠加法：a n =a 1 +（a 2 -a 1 ）+…+（a n -a n-1 ）=1+2+…+n= n(n+1) 2 所以 1 a n = 2 n(n+1) =2（ 1 n - 1 n+1 ） 所以 1 a 1 + 1 a 2 +…+ 1 a 2012 + 1 a 2013 = 2(1- 1 2 + 1 2 - 1 3 +…+ 1 2013 - 1 2014 ) =2× 2013 2014 = 2013 1007 故答案为： 2013 1007