求第九题，幂级数，高等数学
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f(x)
=1/(x-1)
=-1/(1-x)
f(x) = -1/(1-x) => f(0) = -1
f'(x) = -1/(1-x)^2 =>f'(0)/1! = -1
f''(x) =-2/(1-x)^3 =>f''(0)/2! = -1
...
f^(n)(x) = -n!/(1-x)^(n+1) =>f^(n)(0)/n! = -1
1/(x-1) = -1 -x-x^2-...-x^n-.....
=1/(x-1)
=-1/(1-x)
f(x) = -1/(1-x) => f(0) = -1
f'(x) = -1/(1-x)^2 =>f'(0)/1! = -1
f''(x) =-2/(1-x)^3 =>f''(0)/2! = -1
...
f^(n)(x) = -n!/(1-x)^(n+1) =>f^(n)(0)/n! = -1
1/(x-1) = -1 -x-x^2-...-x^n-.....
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