Question

如图，幂级数展开式的求法，是怎么算的，求大神赐教

Answer 1

f(x)=∑<n=0,∞>(n+1)x^n/(n+2)!
= 1/2!+2x/3!+3x^2/4!+4x^3/5!+......+(n+1)x^n/(n+2)!+(n+2)x^(n+1)/(n+3)!+......
f'(x)=2/3!+3*2x/4!+4*3x^2/5!+......, f'(0)=2/3!=1/3;
f''(x)=3!/4!+4*3*2x/5!+......, f''(0)=1/4;
..................................................................................
f^(n)(x)=(n+1)n!/(n+2)!+(n+2)(n+1)!x/(n+3)!+......,
则 f^(n)(0)= 1/(n+2). n=1, 2, 3，......

追问

懂了，谢谢！