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.
I think you might be confusing two different things.
Fact 1: Define $S_n := \sum_{k=1}^n k$. Then $S_n = n(n+1)/2$ for every nonnegative integer $n$.
(You can see this by induction. Certainly the formula is correct for $n = 0$. With $S_{n-1} = (n-1)n/2$, we have $S_{n} = n + S_{n-1} = (n+1)n/2$, as required.)
Then there is the following fact
Fact 2: Define $T_n = \sum_{1 \leq i < j \leq n} 1$. Then $T_n = n(n-1)/2$.
This is just the number of tuples $(k, l)$ with $1 \leq k \leq n$ and $1 \leq l \leq n$ there are with $k < l$. That's just $\binom{n}{2}$, or if you like it's also $(n^2 - n)/2 = n(n-1)/2$.
Best Answer
One way I haven't seen in the answers is to reverse the order of the summation
$$\sum_{i=1}^n\sum_{j=i}^n j+1 = \sum_{j=1}^n\sum_{i=1}^j j+1 = \sum_{j=1}^n j^2 + j $$
which we can use the usual formulas on
$$\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2}$$
This ends up being easier because the summand does not depend on $i$ at all, so it's the equivalent of "integrating" a constant.