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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.

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

Thanks for your help in advance!

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

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  • $\begingroup$ Thanks. But my goal is to keep the summation and change the two terms to only one nested summation. $\endgroup$ – Osama Oct 14 '20 at 12:15
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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. $k+1$).

$$\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{i=1}^{k+1} \sum_{j=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i \\=\sum_{j=1}^{k} \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{j=1}^k\sum_{i=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i+\sum_{i=1}^{k+1} A^{2k+2-(k+1)} B^{k+1-i+(k+1)} C^i \\=\sum_{j=1}^{k}\left( \sum_{i=1}^{j} A^{3k+2-j} B^{1-i+j} C^i + \sum_{i=1}^{k+1} A^{2k+2-j} B^{k+1-i+j} C^i\right)+\sum_{i=1}^{k+1} A^{2k+2-(k+1)} B^{k+1-i+(k+1)} C^i.$$

There is not much more that you can do, though you can probably remove all the summations.

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