Question: Can we get the formula in terms of N and k for this summation series?

$$
A=\sum\_{t=0}^N\sum\_{s=0}^t\sum\_{r=0}^sk^rk^{s-r}k^{t-s}
$$

Answer: 
$$A=\sum\_{t=0}^{N}\sum\_{s=0}^{t}\sum\_{r=0}^{s}k^t=\sum\_{t=0}^N\sum\_{s=0}^t (s+1)k^t = \sum\_{t=0}^N \frac{(t+2)(t+1)}{2}k^t \tag{1}$$
is a partial sum for the Taylor series of $\frac{1}{(1-k)^3}$ in a neighbourhood of $k=0$. 

By multiplying the RHS of $(1)$ by $(1-k)^3$ we get:

> 
> $$ A = \frac{1}{(1-k)^3}\left(1-\binom{N+3}{2}k^{N+1}+(N+3)(N+1)k^{N+2}-\binom{N+2}{2}k^{N+3}\right).\tag{2}$$
> 
> 
>
Another way of determining the last sum is to note that if 

$$S\_N(k)=\frac12 \sum\_{t=0}^N k^{t+2}=\frac{k^2-k^{N+3}}{2(1-k)} \tag 1$$

then upon taking two derivatives reveals that 

$$S\_N''(k)=\sum\_{t=0}^N \frac{(t+2)(t+1)}{2}k^{t}=A \tag 2$$

Thus, taking the second derivative of the right-hand side of $(1)$, we find that

$$S\_N''(k)= \frac12 \left( 2\frac{k^2-k^{N+3}}{(1-k)^3} +2\frac{2k-(N+3)k^{N+2}}{(1-k)^2} +\frac{2-(N+3)(N+2)k^{N+1}}{1-k} \right)$$