Re: [理工] [離散]生成函數

[a016258 (上善若水)]

1/(1+3x)(1+2x)^2

= a/(1+3x) + b/(1+2x) + c/(1+2x)^2

= [ a ( 1+2x)^2 + b * (1+2x)(1+3x) + c(1+3x)  ] / [(1+3x)(1+2x)^2]


=> a ( 1+2x)^2 + b * (1+2x)(1+3x) + c(1+3x) = 1

let x = -1/3

a ( 1 / 9) = 1  => a = 9

let x = -1/2

c ( -1/2) = 1  => c = -2

常數項 a + b + c = 1  => b = -6

有錯還請不吝指正。


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