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函数f(x)=1/(1+x)展开成x-1的幂级数为____．

高老师2年前 (2024-03-26)高等数学（工本）(00023)14

函数f(x)=1/(1+x)展开成x-1的幂级数为____．

∑_n=0^∞(-1)ⁿ•xⁿ(-1﹤x﹤1)。解析：1/(1+x)=1/[1-(-x)]=∑_n=0^∞(-x)ⁿ=∑_n=0^∞(-1)ⁿ•xⁿ(-1﹤x﹤1)

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