将函数f(x)=1⼀X^2+4X+3展开成(X+1)的幂级数并求出收敛区间

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2025-04-12 17:31:59
推荐回答(1个)
回答1:

利用已知级数
1/(1-x) = ∑(n≥1)[x^(n-1)],|x|<1,
可得
f(x) = 1/(x²+4x+3)
= (1/2)[1/(x+1)-1/(x+3)]
= (1/2)[1/(x+1)]-(1/4){1/[1+(x+1)/2]}
= (1/2)[1/(x+1)]-(1/4)∑(n≥1){[-(x+1)/2]^(n-1)},0<|x+1|<1。