I am studying tensors and I have an exercise in finding the number of independent values of a symmetric (2,0)-tensor $B^{IJ}$, where $I, J \in$ {1,2,3, ..., d}. Since the entry of $B^{IJ}$ is the same as $B^{JI}$ my plan was to find a double sum as seen below: $$ \mathop{\sum_{J=1}^{d}\sum_{I=1}^{J}} IJ $$
because $ 1 \le I \le J \le d$. However I don't know how to find the number of terms and not the summation itself.
I know the answer because I asked a friend but I would love to know whether my way of thinking works.
Take care folks!