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I am wondering how to compute this variant of geometric progression: $$ \sum_{i,j \in \{1, \dots, n\}, i \neq j} c^{i+j}? $$

Any help is appreciated!

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3 Answers 3

up vote 4 down vote accepted

$$ \sum_{i,j \in \{1, \dots, n\}, i \neq j} c^{i+j} = \sum_{i \in \{1, \dots, n\}} \sum_{j \in \{1, \dots, n\}\backslash{i}} c^{i}c^{j} \\= \sum_{i \in \{1, \dots, n\}} (\sum_{j \in \{1, \dots, n\}} c^{i}c^{j} - c^{i}c^{i}) \\= \sum_{i \in \{1, \dots, n\}} c^{i}(\sum_{j \in \{1, \dots, n\}} c^{j} - c^{i}) \\= (\sum_{i \in \{1, \dots, n\}} c^{i})^2-\sum_{i \in \{1, \dots, n\}} (c^2)^{i} \\= (c\frac{c^n-1}{c-1})^2-c^2\frac{c^{2n}-1}{c^2-1} $$

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fix an $i$ then compute $$\sum _{i=1}^{n}e^i \left (\frac{e^{n+1}-e}{e-1}-e^{i} \right)$$ Now if you do the multiplication you will get two geometrics sums, which you can compute.

When you fix an $i$ the $j$ take every value expect the fixed $i$ in the numbers $(1,2, \dots ,n)$ so you get a geometric progression minus one term.

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Hint: What if you allowed $i=j$ in the sum, and the subtracted those terms after?

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