Question

求该等式的裂项详细的过程，谢谢！

求该等式的裂项详细的过程，谢谢！请写出具体的裂项过程。我不是很清楚到底是怎么裂这三个式子的……

Answer 1

如下

追问

第二个式子第二项你加一学成减一了……

Answer 2

let

3/[(s+1)(s-2)^2]≡ A/(s+1) + B/(s-2)+C/(s-2)^2
=>
3≡ A(s-2)^2 + B(s+1)(s-2)+C(s+1)
s=2 , =>C=1
s=-1, =>A=1/3
coef. of s^2
A+B=0
1/3+B=0
B=-1/3
3/[(s+1)(s-2)^2]≡ (1/3)[1/(s+1)] -(1/3)[1/(s-2)]+1/(s-2)^2
/
1/(s-2) + 3/[(s+1)(s-2)^2]
=1/(s-2) + (1/3)[1/(s+1)] -(1/3)[1/(s-2)]+1/(s-2)^2