这道题是怎么做？
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f(x) = 1/(1+x) => f(1) =1/2
f'(x) = -1/(1+x)^2 => f'(1)/1! =-1/4
f''(x) = 2/(1+x)^3 => f''(1)/2! =1/8
f'''(x) = -6/(1+x)^4 => f'''(1)/3! =-1/16
f''''(x) = 12/(1+x)^5 => f''''(1)/4! =1/32
f(x)
=1/(1+x)
=1/2 -(1/4)(x-1) +(1/8)(x-1)^2-(1/16)(x-1)^3+(1/32)(x-1)^4+...
f'(x) = -1/(1+x)^2 => f'(1)/1! =-1/4
f''(x) = 2/(1+x)^3 => f''(1)/2! =1/8
f'''(x) = -6/(1+x)^4 => f'''(1)/3! =-1/16
f''''(x) = 12/(1+x)^5 => f''''(1)/4! =1/32
f(x)
=1/(1+x)
=1/2 -(1/4)(x-1) +(1/8)(x-1)^2-(1/16)(x-1)^3+(1/32)(x-1)^4+...
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