Simplifying the sum $\sum\limits_{t=1}^{n} t(t+1)v^t$ For the sum:
$$\sum_{t=1}^{n} t(t+1)v^t,$$
would there be a "simple" formula for this? Such as:
$$S=\sum_{t=1}^{n} tv^t = \frac{\frac{1-v^n}{1-v} - n v^n}{\frac{1}{v} - 1}$$?
(The second sum is the present value of an annuity increasing by \$1 every period, for anyone curious).
I understand that it'll boil down to simplifying the sum to
$$\sum_{t=1}^{n}t^2v^t + S $$
but I'm not sure if there's a nice formula for the first part.
 A: 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 
$$(v^2S_n'
(v))'=2vs_n'(v)+v^2S_n''(v)=\sum_{t=1}^n t(t+1)v^t.$$

For convenient computation, you can differentiate $(1-v)S_n(v)$ twice, giving $$(1-v)S_n'(v)-S_n(v)$$ and $$(1-v)S_n''(v)-2S_n'(v).$$
A: 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 $S_n$.
A: Let $S = \sum_{t = 1}^{n} t^2 v^t$
$S =  v + 4v^2 + 9v^3 + ... + n^2 v^n$
$Sv = \ \ \ \ \ v^2 +  4v^3 + ... + n^2 v^{n+1}$
$S(1-v) = v + 3v^2 + 5v^3 + ... (n^2 - (n-1)^2)v^n - n^2 v^{n+1}$
$S(1-v)v = \ \ \ \ \ v^2 + 3v^3 + ... ((n-1)^2 - (n-2)^2)v^n + (n^2 - (n-1)^2)v^{n+1} - n^2 v^{n+2}$
$S(1-v)^2 = v + 2v^2 + 2v^3 + ... 2v^n - (n^2 + 2n - 1)v^{n+1} + n^2 v^{n+2}$
$S(1-v)^2 = v + 2v^2(1 + v + v^2 + ... v^{n-2}) - (n^2 + 2n - 1)v^{n+1} + n^2 v^{n+2}$
$S = \frac{v}{(1-v)^2} - \frac{2v^2 (v^{n-1} - 1)}{(1-v)^3} - \frac{(n^2 + 2n -1 )v^{n+1}}{(1-v)^2} + \frac{n^2v^{n+2}}{(1-v)^2}$
EDIT: This can be further simplified to:
$S = -\frac{n^2v^{n+1}}{(1-v)} - \frac{2nv^{n+1}}{(1-v)^2} + \frac{v(1-v^n)(1+v)}{(1-v)^3}$
Verify: $n = 0$, gives $S = 0$
$n = 1$; $S = -\frac{v^2}{(1-v)} - \frac{2v^2}{(1-v)^2} + \frac{v(1-v)(1+v)}{(1-v)^3} = v$
