Evaluate $\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]$ at $t=1$ I need to find a "nice" formula for the evaluation of 
$\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]$ at t=1, where $d_j \in \mathbb{N}$.
I have already proved that 
\begin{equation}
\frac{\partial}{\partial t} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]_{| t=1} =\frac{1}{2} \cdot \prod_{i=1}^k d_i \cdot \left[ \sum_{i=1}^k d_i -k \right]
\end{equation}
and now I'm looking for a similar concise formula for the second derivative at t=1.
 A: Let $$f_i(x)=\sum_{n=0}^{d_i-1}x^n=1+x+\cdots+x^{d_i-1}$$ and let $$\Psi=\prod_{j=1}^kf_j(t)$$
We have that $$\Phi=\frac\partial{\partial t}\Psi=\Psi\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)$$ Therefore $$\frac{\partial^2}{\partial t^2}\Psi=\frac\partial{\partial t}\Phi=\Psi'\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)+\Psi\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)=\Psi\left(\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)^2+\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)'\right)$$
Finally after simplification
$$\Psi\left(\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)^2+\sum^k_{j=1}\frac{f_j(t)f_j''(t)-f_j'(t)^2}{f_j(t)^2}\right)$$
The rest is just computations of the derivative at of $f_j$ at $t=1$ which should simplify nicely in terms of $d_j$.

Further Calculations
$$f_j(1)=d_j$$
$$f_j'(1)=\frac{d_j(d_j-1)}{2}$$
$$f_j''(1)=\frac{d_j(d_j-1)(d_j-2)}{3}$$
Thus you may finally plug and chug.

After the plug and chug I get $$\frac{\partial^2}{\partial t^2}\Psi=\left(\prod^k_{j=1}d_j\right)\left(\left(\sum^k_{j=1}\frac{d_j-1}2\right)^2-\sum^k_{j=1}\frac{(d_j-1)(d_j-5)}{2}\right)$$
A: This is basically the same calculation as Alitzer gives, but more readily generalized. The Leibniz rule (generalized to multiple derivatives & factors 
[1]) states that 
$$\bigg(\prod_{j=1}^k f_j\bigg)^{\!(n)}=\sum_{\substack{c_1,\ldots,c_k\ge 0\\ c_1+\cdots+c_m=n}}\binom{n}{c_1,\cdots, c_k}f_1^{(c_1)}\cdots f_k^{(c_k)}$$ where $\binom{n}{c_1,\cdots, c_k}=n!/(c_1!\cdots c_k!)$ is a multinomial coefficient.
For the case of $n=2$ and $f_j=\sum_{k=0}^{d_j-1}t^k$ as above, this gives 
\begin{align}
\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]
&=\sum_{c_1+\cdots +c_m=2}\binom{2}{c_1,\cdots,c_m}f_1^{(c_1)}\cdots f_k^{(c_k)}
\end{align}
Note that there are only two kinds of terms that show up in this sum: either exactly one of the $c's$ is 2, or some pair of them is 1 (with all others zero). So this simplifies to 
$$\sum_{i=1}^k\left( f_1\cdots f_i^{(2)}\cdots f_k\right)+\sum_{i<j}^k\left( f_1\cdots f_i^{(1)}\cdots f_j^{(1)}\cdots f_k\right)$$
which we now need to evaluate at $t=1$. Then
$$f_j|_{t=1}=d_j,
\hspace{.5cm} f^{(1)}_j|_{t=1}=\sum_{k=0}^{d_j-1} k = \frac{d_j(d_j-1)}{2},\\
\hspace{.5cm} f^{(2)}_j|_{t=1}=\sum_{k=0}^{d_j-1} k(k-1) = \frac{d_j(d_j-1)(d_j-2)}{3}.$$
Plugging these in and factoring a bit gives the final result as $$\left(\prod_{j=1}^m d_j\right)\left[\sum_{i=1}^{k}\frac{(d_i-1)(d_i-2)}{3}
+\sum_{i<j}\frac{(d_i-1)(d_j-1)}{4}\right]$$
A: I've cheated, I admit:
$$\frac{\partial ^2\left(\prod_j^k \sum _{d=0}^j t^d\right)}{\partial t^2}$$
at $t=1$ seems to equal  
$$\frac{1}{144} (k-1) k (k+1) (9 k+22) (k+1)!$$
I can show you the cheat, but cannot provide a proof.
