I should state that the following steps depend on the formula for a geometric series:
$$\sum_{j=1}^{k} r^{j-1} = \frac{1-r^k}{1-r}$$
The sum in question is
$$p_1 p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} \sum_{i=1}^{k-j} (1-p_1)^{i-1} $$
Evaluate the inner sum first:
$$\sum_{i=1}^{k-j} (1-p_1)^{i-1} = \frac{ 1 - (1-p_1)^{k-j} }{p_1} $$
Now the sum is
$$\begin{align} p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} \left [ 1 - (1-p_1)^{k-j} \right ] &= 1 - (1-p_2)^{k-1} - p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} (1-p_1)^{k-j} \\ \end{align} $$
The sum on the right-hand side may be evaluated as follows:
$$\begin{align}\sum_{j=1}^{k-1} (1-p_2)^{j-1} (1-p_1)^{k-j} &= (1-p_1)^{k-1} \sum_{j=1}^{k-1} \left ( \frac{1-p_2}{1-p_1} \right )^{j-1} \\ &= (1-p_1)^{k-1} \frac{ 1 - \left ( \frac{1-p_2}{1-p_1} \right )^{k-1}}{1 - \frac{1-p_2}{1-p_1}} \\ &= (1-p_1)\frac{(1-p_1)^{k-1} - (1-p_2)^{k-1}}{p_2-p_1} \\ \end{align} $$
Now we can put this all together:
$$\begin{align} 1 - (1-p_2)^{k-1} - p_2 (1-p_1)\frac{(1-p_1)^{k-1} - (1-p_2)^{k-1}}{p_2-p_1} \\ = 1 - \frac{(p_2-p_1)(1-p_2)^{k-1} + p_2 (1-p_1)[(1-p_1)^{k-1} - (1-p_2)^{k-1}]}{p_2-p_1} \\ \end{align}$$
which, after some cancellation and consolidation, produces the following result for the sum:
$$1 - \frac{p_1 (1-p_2)^k - p_2 (1-p_1)^k}{p_1-p_2}$$
and is equal to the result stated above.
$$
\begin{align}
\frac{d}{d\mu} \sum_{i=1}^v \sum_{t=1}^{r_i} (y_{it}-\mu - \tau_i)^2
& = \sum_{i=1}^v \sum_{t=1}^{r_i} \frac{d}{d\mu} (y_{it}-\mu - \tau_i)^2 \\[6pt]
& = \sum_{i=1}^v \sum_{t=1}^{r_i} 2(y_{it}-\mu - \tau_i)(-1) \\[6pt]
& = -2 \left(\sum_{i=1}^v \sum_{t=1}^{r_i} y_{it}\right) -2 \left(\sum_{i=1}^v \sum_{t=1}^{r_i} \mu \right) -2 \left(\sum_{i=1}^v \sum_{t=1}^{r_i} \tau_i\right) \\[6pt]
& = -2 y_{\cdot\cdot} - n\mu + \sum_{i=1}^v r_i t_i.
\end{align}
$$
The fact that $\displaystyle\sum_{i=1}^v \sum_{t=1}^{r_i} y_{it}=y_{\cdot\cdot}$ is just the definition of $y_{\cdot\cdot}$. The second sum is $n\mu$ because it's a sum of $n$ terms, each equal to $\mu$. For the third sum, one has $\displaystyle\sum_{t=1}^{r_i} \tau_i = r_i\tau_i$ because it's a sum of $r_i$ terms, each equal to $\tau_i$.
PS: It would be a mistake to write something like this:
$$\frac{d}{d\tau_i}\sum_{i=1}^v \sum_{t=1}^{r_i} (y_{it}-\mu - \tau_i)^2$$
Let's say $n=3$, so that $i=\text{either 1, 2, or 3}$. In the expression $\dfrac{d}{d\tau_i}$, the $i$ is either $1$, $2$, or $3$. But in the expression $\displaystyle\sum_{i=1}^3$, the index $i$ runs through the whole list of three values and you add up the terms, while the $i$ in $\dfrac{d}{d\tau_i}$ stays put!
So instead, write
$$\frac{d}{d\tau_i}\sum_{j=1}^v \sum_{t=1}^{r_j} (y_{jt}-\mu - \tau_j)^2.$$
This is
$$\sum_{j=1}^v \frac{d}{d\tau_i} \sum_{t=1}^{r_j} (y_{jt}-\mu - \tau_j)^2,$$
and
$$
\frac{d}{d\tau_i} (y_{jt}-\mu - \tau_j)^2 = \begin{cases} 0 & \text{when }j\ne i, \\[8pt]
\sum_{t=1}^{r_i} 2(y_{it} -\mu -\tau_i)(-1) & \text{when }j=i. \end{cases}
$$
Thus all of the terms in the sum $\sum_{j=1}^v$ are $0$ except the $i$th term.
Best Answer
This sort of thing usually comes up in the context of trying to interchange the derivative with the sum: that is, you would like to have
$$\frac{d}{dx} \sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty f'_n(x)$$
by analogy with the case of finite summation. This interchange can sometimes fail. The most basic criterion that I have heard of is one of the "advanced calculus criteria" (so called because it is taught in undergraduate real analysis and has no famous name). It requires that $\sum_{n=0}^\infty f'_n(x)$ converges uniformly in $x$ (over the domain on which you have the equality). Milder criteria exist; for instance there is a variant based on the dominated convergence theorem from measure theory.
I'm not sure if this answers your question because I'm not sure what "sound" means in this context.