How to proceed further in this Arithmetico-Geometric Progression problem
Your approach is correct! Now for $x=\frac{4n+1}{4n-3}$, we find that $$S_n=\frac{x^n((4n-3)x-(4n+1))+(3x-1)}{(x-1)^2}=\frac{0+(3x-1)}{(x-1)^2}=n(4n-3).$$
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 $$s_n=n(4n-3)$$