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.
It will help to properly indent your code:
int sum = 0;
for (int i = 1; i <= 4; i++)
for (int j = 1; j <= 3; j++)
sum += i * j;
You can work outside-in, or inside-out. Let's do it inside-out. First, then, a sum for the inner for loop, with index $j$ and parameter $i$:
$$
S_i = \sum_{j=1}^3 i*j
$$
Now the outer loop, with index $i$, summing over the $S_i$:
$$
\begin{align}
sum &= \sum_{i=1}^4 S_i \\
sum &= \sum_{i=1}^4 \sum_{j=1}^3 i*j \\
\end{align}
$$
Best Answer
$$\sum^n_{i = 1} \sum^n_{j = i+1} j - i + 2= \sum^n_{i = 1} \sum^{n-i}_{j =1} j + 2=\sum^n_{i = 1}\frac{(n-i)^2+(n-i)}{2}+2(n-i)$$ $$=\sum_{i=1}^{n-1}\frac{i^2+i}{2}+2i=\sum_{i=1}^{n-1}\frac{i^2+5i}{2}=\frac{1}{2}\sum_{i=1}^{n-1}i^2+\frac{5}{2}\sum_{i=1}^{n-1}i=\frac{n(n-1)(n+7)}{6}$$