sequence proof
Let y=1+x or x=y−1
f(x)=1y+2y2+3y3+4y4+…+n−1yn−1+nyn|⋅yyf(x)=1+2y+3y2+4y3+5y4+…+nyn−1yf(x)−f(x)=1+1y+1y2+1y3+1y4+…+1yn−1−nyn(y−1)f(x)=1+1y+1y2+1y3+1y4+…+1yn−1−nyn|y−1=xxf(x)=1+1y+1y2+1y3+1y4+…+1yn−1−nyn
xf(x)=1+1y+1y2+1y3+1y4+…+1yn−1⏟=s (sum of a geometric progression)−nynxf(x)=s−nyns=1+1y+1y2+1y3+1y4+…+1yn−1|:ysy=1y+1y2+1y3+1y4+…+1yn−1+1yns−sy=1−1yns(1−1y)=1−1yns=1−1yn1−1yxf(x)=1−1yn1−1y−nyn|y=1+xxf(x)=1−1(1+x)n1−11+x−n(1+x)n
xf(x)=1−1(1+x)n1−11+x−n(1+x)nxf(x)=(1+x)n−1(1+x)n1+x−11+x−n(1+x)nxf(x)=(1+x)n−1(1+x)nx1+x−n(1+x)nxf(x)=((1+x)n−1)(1+x)x(1+x)n−n(1+x)n⋅xxxf(x)=((1+x)n−1)(1+x)−nxx(1+x)n|:xf(x)=((1+x)n−1)(1+x)−nxx2(1+x)nf(x)=(1+x)n+1−(1+x)−nxx2(1+x)n