# Can this two nested sums be reduced to one nested sum?

I have encountered this sum in my homework. I want to simplify or reduce the following sum to only one nested sum, where k is a positive integer.

I think it is possible to make the second nestes sum term contain the first nested sum. But I don't know how.

$$\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{i=1}^{k+1} \sum_{j=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i$$

• The second sum is the product of two geometric series. – Empy2 Oct 14 '20 at 11:54

The main idea is to separate different powers with respect to $$i$$ and $$j$$ and leave only things that depend on $$i$$ under the sum with respect to $$i$$.
For example, for your first term, you get $$\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i = A^{3k+2}B\sum_{j=1}^{k} A^{-j}B^{j}\sum_{i=1}^{j} B^{-i} C^i.$$ You can easily find the inner sum (geometric progression, basically), then the outer sum is done likewise.
You can swap the two summations on the right, and regroup under a single summation on $$j$$. But beware that the final indexes differ ($$k$$ vs. $$k+1$$).
$$\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{i=1}^{k+1} \sum_{j=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i \\=\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{j=1}^k\sum_{i=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i+\sum_{i=1}^{k+1} A^{2k+2-(k+1)} B^{k+1-i+(k+1)} C^i \\=\sum_{j=1}^{k}\left( \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{i=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i\right)+\sum_{i=1}^{k+1} A^{2k+2-(k+1)} B^{k+1-i+(k+1)} C^i.$$