第1个回答 2016-04-19
详细答案在图片上，希望得到采纳，谢谢≧◔◡◔≦
第2个回答 2016-04-19
(1)
f(x) = 1/(1-x)^3
f'(x) = 3/(1-x)^4
f^(n)(x) = (n+2)!/[2(1-x)^(n+3) ]
f^(n)(0) =(n+2)!/2
f^(n)(0)/n! = (n+1)(n+2)/2
f(x) = f(0) +[f'(0)/1!]x+...+[f^(n)(0)/1!]x^n +...
=1 + 3x+6x^2+...+[(n+1)(n+2)/2]x^n+...
(2)
f(x) =(a+x)ln(1+x) =>f(0) = 0
f'(x) = (a+x)/(1+x) + ln(1+x) =>f'(0)/1! = a
= 1 + (a-1)/(1+x) +ln(1+x)
f''(x) = -(a-1)/(1+x)^2 -1/(1+x) =>f''(0)/2!= -a/2
f'''(x) = 2(a-1)/(1+x)^3 + 1/(1+x)^2 =>f'''(0)/3! =(2a-1)/6
for n >=2
f^(n)(x) = (-1)^(n-1) . [ (a-1). (n-1)!/(1+x)^n + (n-2)!/(1+x)^(n-1)]
f^(n)(0) =(-1)^(n-1) . [ (a-1). (n-1)! + (n-2)!]
=(-1)^(n-1) . (n-2)! .[ (n-1)(a-1)+1]
f^(n)(0)/n! =(-1)^(n-1) [ (n-1)(a-1)+1]/[n(n-1) ]
f(x)
=ax - (a/2)x^2+[(2a-1)/6]x^2 +...+(-1)^(n-1).[(n-1)(a-1)+1]/[n(n-1) ]x^n +...本回答被网友采纳
f(x) = 1/(1-x)^3
f'(x) = 3/(1-x)^4
f^(n)(x) = (n+2)!/[2(1-x)^(n+3) ]
f^(n)(0) =(n+2)!/2
f^(n)(0)/n! = (n+1)(n+2)/2
f(x) = f(0) +[f'(0)/1!]x+...+[f^(n)(0)/1!]x^n +...
=1 + 3x+6x^2+...+[(n+1)(n+2)/2]x^n+...
(2)
f(x) =(a+x)ln(1+x) =>f(0) = 0
f'(x) = (a+x)/(1+x) + ln(1+x) =>f'(0)/1! = a
= 1 + (a-1)/(1+x) +ln(1+x)
f''(x) = -(a-1)/(1+x)^2 -1/(1+x) =>f''(0)/2!= -a/2
f'''(x) = 2(a-1)/(1+x)^3 + 1/(1+x)^2 =>f'''(0)/3! =(2a-1)/6
for n >=2
f^(n)(x) = (-1)^(n-1) . [ (a-1). (n-1)!/(1+x)^n + (n-2)!/(1+x)^(n-1)]
f^(n)(0) =(-1)^(n-1) . [ (a-1). (n-1)! + (n-2)!]
=(-1)^(n-1) . (n-2)! .[ (n-1)(a-1)+1]
f^(n)(0)/n! =(-1)^(n-1) [ (n-1)(a-1)+1]/[n(n-1) ]
f(x)
=ax - (a/2)x^2+[(2a-1)/6]x^2 +...+(-1)^(n-1).[(n-1)(a-1)+1]/[n(n-1) ]x^n +...本回答被网友采纳
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