f(x) = 1/(2+x) =>f(0) =1/2
f'(x) = -1/(2+x)^2 =>f'(0)/1! =-1/4
f''(x) = 2/(2+x)^3 =>f''(0)/2! =1/8
...
f^(n)(x)= (-1)^n.n!/(2+x)^(n+1) =>f^(n)(0)/n! =(-1)^n/2^(n+1)
1/(2+x)
=f(x)
=f(0) + [f'(0)/1!]x+[f''(0)/2!]x^2+...+[f^(n)(0)/n!]x^n+...
=1/2 - (1/4)x +(1/8)x^3+...+ [(-1)^n/2^(n+1) ]x^n+...
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