Question

15题怎么做？答案是什么？？急！！

Answer 1

设 1/n(n+1)(n+2)=A/n+B/(n+1)+C/(n+2) [ABC是常数]

则 通分 得 上式=[(A+B+C)n^2+(5A+4B+3C)n+(6A+3B+2C)]n(n+1)(n+2)

又上式=1/n(n+1)(n+2) 所以 A+B+C=0 5A+4B+3C=0 6A+3B+2C=1

得 A=0.5 B=-1 C=0.5

所以 1/n(n+1)(n+2) = 0.5[1/n+1/(n+2)-2/(n+1)]
an=0.5[1/n+1/(n+2)-2/(n+1)]+1/（n+1）
=0.5[1/n+1/(n+2)]

自己带入算吧