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the solution sheet of an exercice I was working on says in particular:

$$ \sum_{k=0}^{n}\sum_{i=0}^{k}\binom{k}{i}t^ia^{k-i}\alpha_k = \sum_{i=0}^{n}t^i\sum_{k=i}^{n}\binom{k}{i}a^{k-i}\alpha_k $$

I can convince myself of this by spelling out summations on each side, and I think this could probably be proven by induction. But my question is rather: how does one see (or intuit) this relation starting with the LHS? Is there a specific method to be mindful of here?

Thanks

EDIT: In the meantime I found [this thread][1] in a similar vein, whose ticked answer parallels DavidW's here.

Mogu
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3 Answers3

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Intuitively best seen by recognizing the double sum as single sum $$\sum_{(i,k)\text{ with }0\le i\le k\le n} $$

  • Interesting! I can see that this works for le LHS, but having a harder time with the RHS. I'll try to work this out. Thanks – Mogu Oct 22 '20 at 10:48
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Imagine the terms in a matrix indexed by $k$ for rows and $i$ for columns. The right side finds the total by summing along the rows instead of down the columns.

DavidW
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More generally$$\sum_{k=0}^n\sum_{i=0}^ka_{ki}b_ic_{ki}=\sum_{k=0}^n\sum_{i=0}^kb_ia_{ki}c_{ki}=\sum_{0\le i\le k\le n}b_ia_{ki}c_{ki}=\sum_{i=0}^nb_i\sum_{k=i}^na_{ki}c_{ki}.$$

J.G.
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  • That is akin to what Hagen was saying. But ok, it might be worth proving it once and to keep the pattern in mind. This is less about intuition, but that does adress the method aspect of my question! – Mogu Oct 22 '20 at 10:53