1+2q+3q2+...+nqn−1=1−(n+1)qn+nqn+1(1−q)21+2q+3q2+...+nqn−1=1−(n+1)qn+nqn+1(1−q)2

Hypothesis:

 

F(x)=1+2q+3q2+...+xqx−1=1−(x+1)qx+xqx+1(1−q)2F(x)=1+2q+3q2+...+xqx−1=1−(x+1)qx+xqx+1(1−q)2

Proof:

 

P1|F(x)=1−(x+1)qx+xqx+1(1−q)2+(x+1)qx=1−(x+2)qx+1+xqx+2(1−q)2P1|F(x)=1−(x+1)qx+xqx+1(1−q)2+(x+1)qx=1−(x+2)qx+1+xqx+2(1−q)2
P2|1−(x+1)qx+xqx+1+[(x+1)(1−q)2]qx(1−q)2=1−(x+2)qx+1+xqx+2(1−q)2P2|1−(x+1)qx+xqx+1+[(x+1)(1−q)2]qx(1−q)2=1−(x+2)qx+1+xqx+2(1−q)2
P3|xqx+1+[−(x+1)]qx+1+[(x+1)(1−q)2]qx(1−q)2=xqx+2−(x+2)qx+1+1(1−q)2|P3|xqx+1+[−(x+1)]qx+1+[(x+1)(1−q)2]qx(1−q)2=xqx+2−(x+2)qx+1+1(1−q)2|

Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' qxqx are added next.

 

P4|xqx+1+[−(x+1)+(x+1)(<12q0+(21)1q−10q2>)]qx+1(1−q)2=xqx+2−(x+2)qx+1+1(1−q)2P4|xqx+1+[−(x+1)+(x+1)(<12q0+(21)1q−10q2>)]qx+1(1−q)2=xqx+2−(x+2)qx+1+1(1−q)2
P5|xqx+1+[2xq−xq2+2q−q2]qx+1(1−q)2=xqx+2−(x+2)qx+1+1(1−q)2P5|xqx+1+[2xq−xq2+2q−q2]qx+1(1−q)2=xqx+2−(x+2)qx+1+1(1−q)2

I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated.

Thanks in advance.