TITLE: How to proceed further in this Arithmetico-Geometric Progression problem
QUESTION [6 upvotes]: Question:
The sum to $n$ terms of the series,
$S=1+5(\frac{4n+1}{4n-3})+9(\frac{4n+1}{4n-3})^2+13(\frac{4n+1}{4n-3})^3+....$

The following image is my approach towards the problem.

Could you please tell how to proceed? I proceeded in this way as I did not want to mug-up the complicated formula given for AGP summation.
The final answer is $n(4n-3)$

REPLY [3 votes]: Note that $$x=1+\frac {4}{4n-3}$$ so $$1-x=\frac {-4}{4n-3}$$ 
Substitute in  your last line and you will get the $$s_n=n(4n-3)$$