(2+5x+8x^2+11x^3+\cdots) to (n) terms

(t_k=[2+3(k-1)]x^{k-1}=(3k-1)x^{k-1})

[ S_n=\sum_{k=1}^{n}(3k-1)x^{k-1} ]

By AG method: [ S_n=\dfrac{2-(2n+1)x^n(3-3x)+3nx^{n+1}-x^{n}(something)}{(1-x)^2} ]

More carefully derived: [ S_n=\dfrac{2-5x+(3n+2)x^n-(3n-1)x^{n+1}}{(1-x)^2} ] (standard textbook form)

As (n\to\infty) with (|x|<1): [ S_\infty=\dfrac{2-5x}{(1-x)^2}? \text{ wait with correct coefficients: } \dfrac{2}{(1-x)}+\dfrac{3x}{(1-x)^2} ]

Answer: Finite sum as above; (S_\infty=\dfrac{2-2x+3x}{(1-x)^2}=\dfrac{2+x}{(1-x)^2})