consider
1/(1-x) = 1+x+x^2+x^3+.
[1/(1-x)]' =[ 1+x+x^2+x^3+.]'
1/(1-x)^2 = 1+2x+3x^2+.
x/(1-x)^2 = x+2x^2+3x^3+.
also
x/(1-x)^2 = x+2x^2+3x^3+.
[x/(1-x)^2]' = [x+2x^2+3x^3+.]'
[(1-x)^2+2x(1-x)] /(1-x)^4 = 1+(2^2)x+(3^3)x^2+...
(1+x)/((1-x)^3 =1+(2^2)x+(3^3)x^2+...
x(1+x)/((1-x)^3 = x+(2^2)x^2+(3^3)x^3+...
I= summation(n:1->无穷)(n^2+n)x^n
= (x+(2^2)x^2+...)+(x+2x^2+3x^3+..)
= x(1+x)/((1-x)^3 + x/(1-x)^2
= [x(1+x) + x(1-x)]/(1-x)^3
=2x/(1-x)^3 consider
1/(1-x) = 1+x+x^2+x^3+.
[1/(1-x)]' =[ 1+x+x^2+x^3+.]'
1/(1-x)^2 = 1+2x+3x^2+.
x/(1-x)^2 = x+2x^2+3x^3+.
also
x/(1-x)^2 = x+2x^2+3x^3+.
[x/(1-x)^2]' = [x+2x^2+3x^3+.]'
[(1-x)^2+2x(1-x)] /(1-x)^4 = 1+(2^2)x+(3^3)x^2+...
(1+x)/((1-x)^3 =1+(2^2)x+(3^3)x^2+...
x(1+x)/((1-x)^3 = x+(2^2)x^2+(3^3)x^3+...
I= summation(n:1->无穷)(n^2+n)x^n
= (x+(2^2)x^2+...)+(x+2x^2+3x^3+..)
= x(1+x)/((1-x)^3 + x/(1-x)^2
= [x(1+x) + x(1-x)]/(1-x)^3
=2x/(1-x)^3