将函数f（x）=1x2+4x+3展开成（x-1）的幂级数．

Question

将函数f（x）=(1)/(x2+4x+3)展开成（x-1）的幂级数．

Answer 1

因为f(x)＝(1)/(x2+4x+3)＝(1)/((x+1)(x+3))＝(1)/(2(1+x))−(1)/(2(3+x))＝(1)/(4(1+(x−1)/(2)))−(1)/(8(1+(x−1)/(4)))，
又因为 (1)/(1+x)=

∞

n＝0

(−1)nxn，-1＜x＜1，
故在−1＜x＜3中，(1)/(4(1+(x−1)/(2)))＝(1)/(4)

∞

n＝0

((−1)n)/(2n)(x−1)n，
在−3＜x＜5中，(1)/(8(1+(x−1)/(4)))＝(1)/(8)

∞

n＝0

((−1)n)/(4n)(x−1)n，
再注意到（-1，3）∩（-3，5）=（-1，3），
因此，f(x)＝(1)/(x2+4x+3)＝

∞

n＝0

(−1)n((1)/(2n+2)−(1)/(22n+3))(x−1)n，（-1＜x＜3）．