Simplifying the sum $\sum\limits_{t=1}^{n} t(t+1)v^t$

Let $S_n=\sum_{t=1}^n t(t+1)x^t$. Then $$(1-x)S_n=\sum_{t=1}^n \left[t(t+1)-(t-1)t\right]x^t-n(n+1)x^{n+1} =2\sum_{t=1}^ntx^t-n(n+1)x^{n+1}.$$ So if you can work out $\sum_{t=1}^ntx^t$ you can find $S_n$.

Hint:

Let

$$S_n(v)=\sum_{t=1}^n v^t=v\frac{1-v^n}{1-v}.$$

Then

$$v^2S_n'(v)=\sum_{t=1}^n tv^{t+1}$$

and

$$(v^2S_n' (v))'=2vs_n'(v)+v^2S_n''(v)=\sum_{t=1}^n t(t+1)v^t.$$

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For convenient computation, you can differentiate $(1-v)S_n(v)$ twice, giving $$(1-v)S_n'(v)-S_n(v)$$ and $$(1-v)S_n''(v)-2S_n'(v).$$