Let n=5:
1+(1+x)+(1+x+x2)+(1+x+x2+x3)+(1+x+x2+x3+x4),
(1-x)/(1-x)+(1-x2)/(1-x)+(1-x3)/(1-x)+(1-x4)/(1-x)+(1-x5)/(1-x)=
(1/(1-x))[5-x-x2-x3-x4-x5]=(1/(1-x))[5-x(1+x+x2+x3+x4)]=
(1/(1-x))[5-x(1-x5)]/(1-x)]=(1/(1-x)2)[5-5x-x(1-x5)]=(1/(1-x)2)(5-6x+x5)
Generalising: Sn=(1/(1-x)2)[n-nx-x(1-xn)], or (1/(1-x)2)(n-(n+1)x+xn+1).
When n=1, S1=(1/(1-x)2)[1-2x+x2]=(1/(1-x)2)(1-x)2=1.
When n=2, S2=(1/(1-x)2)(2-3x+x3)=(1/(1-x)2)[(x+2)(x-1)2]=x+2=1+(1+x).
These examples confirm the rule.