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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?

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  • $\begingroup$ 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. $\endgroup$ – Ove Ahlman Oct 28 '15 at 9:27
  • $\begingroup$ @OveAhlman I have laid my thoughts and approach on solving this problem as you said. Any help would be appreciated. $\endgroup$ – Anish Sharma Oct 28 '15 at 9:48
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Hints:

  • 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 $\sum_{r=2}^{\infty}y^r$.

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