# Solution to closed form of a Generating function

Can anyone give me the closed form of the generating function $$\Sigma ^\infty_{r=2} r.4^r.x^r$$ I am trying to solve recurrence relation using generating functions and this is one of the terms. I need to write its closed form. Explanation would be helpful. For example, we know $$\Sigma_{n=2}^\infty 2^{n-2}.x^n= x^2+2.x^3+2^2.x^4+....$$ $$=x^2(1+2x+(2x)^2+...)$$$$=x^2.(1-2x)^{-1}$$ Similarly, I am approaching this problem as$$\Sigma ^\infty_{r=2} r.4^r.x^r=2.4^2.x^2+3.4^3.x^3+....$$$$=4^2.x^2(2+3.4x+4.4^2.x^2+54^3.x^3+....)$$$$=4^2.x^2(2+3.4x+4.(4x)^2+5.(4x)^3+....)$$ From here, what should I do? can anyone help me with this approach?

• What are your thoughts on how to solve this problem? It is easier for us to answer you with a solution which can help you if you provide some insight into your toolbox to solving this. – Ove Ahlman Oct 28 '15 at 9:27
• @OveAhlman I have laid my thoughts and approach on solving this problem as you said. Any help would be appreciated. – Anish Sharma Oct 28 '15 at 9:48

• substituting $y=4x$ write the summation as: $y\sum_{r=2}^{\infty}ry^{r-1}$
• differentiate function $\sum_{r=2}^{\infty}y^r$ and compare.
• find a closed form for $\sum_{r=2}^{\infty}y^r$.