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.
Best Answer
1) Looks fine to me.
2) Notice this:
$$\begin{array}{}\displaystyle\sum_{x=1}^n\sum_{y=1}^n(x+y-1)&\displaystyle\vphantom{\cfrac11}=\sum_{x=1}^n\sum_{y=1}^nx&\displaystyle+\sum_{x=1}^n\sum_{y=1}^ny&\displaystyle-\sum_{x=1}^n\sum_{y=1}^n1\\\vphantom{\cfrac11}&\displaystyle=\sum_{x=1}^nx\sum_{y=1}^n1&\displaystyle+\sum_{x=1}^n1\sum_{y=1}^ny&\displaystyle-\sum_{x=1}^n1\sum_{y=1}^n1\\\vphantom{\cfrac11}&\displaystyle=\frac{n(n+1)}2\cdot n&\displaystyle+n\cdot\frac{n(n+1)}2&\displaystyle-n\cdot n\\&=n^3\end{array}$$