Question

求幂级数（如下图）的和函数。（解答过程）

Answer 1

解：令上述幂级数的和函数等于E
E=1+(2^2)x+(3^2)x^2+(4^2)x^3+…………+[(n+1)^2]x^n
xE=x+(2^2)x^2+(3^2)x^3+(4^2)x^4+…………+[(n+1)^2]x^(n+1)
两式相减得：
(1-x)E=1-[(n+1)^2]x^(n+1)+(2^2-1)x+(3^2-2^2)x^2+(4^2-3^2)x^3+…………+[(n+1)^2-n^2]x^n
=[1+3x+5x^2+7x^3+9x^4+…………+(2n+1)x^n]-[(n+1)^2]x^(n+1)
x(1-x)E=[x+3x^2+5x^3+7x^4+…………+(2n-1)x^n+(2n+1)x^(n+1)]-[(n+1)^2]x^(n+2)
两式又相减可得：
(1-x)E - x(1-x)E=(x-1)^2E
=1 - [(n+1)^2]x^(n+1) - (2n+1)x^(n+1) + [(n+1)^2]x^(n+2) +2[x+x^2+x^3+x^4+……+x^n]
=1 - [(n+1)^2]x^(n+1) - (2n+1)x^(n+1) + [(n+1)^2]x^(n+2) + 2[x(1-x)^n]/(1-x)
=[x+1-2.x^(n+1)]/(1-x) + [(n+1)^2]x^(n+2) - (n^2+4n+2)x^(n+1) = (x-1)^2.E
所以E为：
E={[x+1-2.x^(n+1)]/(1-x) + [(n+1)^2]x^(n+2) - (n^2+4n+2)x^(n+1)} / [(x-1)^2]
=[(2.x^(n+1) - (x+1)] /(x-1)^3 + [(n+1)^2.x^(n+2)] /(x-1)^2 - [(n^2+4n+2).x^(n+1)] /(x-1)^2

终于输完了！就是这样的了，希望对楼主有帮助！

Answer 2

虽然楼上输的很辛苦...
用这个符号表示求和符号∑(n,0,+inf)
对上式逐项积分得到∑(n,0,+inf)(n+1)x^(n+1)
设S=∑(n,0,+inf)(n+1)x^(n+1)=x+2x^2+3x^3++4x^4....
xS=x^2+2x^3+3x^4....
(1-x)S=x+x^2+x^3+.....
由泰勒展开知1/(1-x)=1+x+x^2+...
所以(1-x)S=x(1+x+x^2+..)=x/(1-x)
S=x/(1-x)^2
然后对S求导就得到所求的和式
S‘=(x+1)/(1-x)^3
若满意,请采纳！.....