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$$ =\sum_{i=1}^{n} \frac{(n-i)(n-i+1)}{2} $$$$ =\sum_{i^{\prime}=0}^{n-1} \frac{i^{\prime}\left(i^{\prime}+1\right)}{2} $$

I see that $i' = n - i$, but shouldn't then the bounds be different? Like for the lower bound $n - 1$ as you substitute in $1$ for $i$ and vice versa?

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Hilberto1
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  • The bounds changed from $1, n$ to $n-1, n-n$. – player3236 Oct 01 '20 at 16:49
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    Just set $i'=n-i$. – TheSilverDoe Oct 01 '20 at 16:49
  • "shouldn't the bounds be different? like for the lower bound n-1..." If I wanted to add $a+b+c+d$ I could do so from left to right, starting with $a$, then adding $b$, then adding $c$, etc... or, and stay with on me on this, I could have added from right to left! So, start with $d$ and then add $c$ and then add $b$ and finally add $a$. That's all that happened here. Instead of adding from left to right we just changed to adding from right to left and adjusted the sum and indexing variable to match. – JMoravitz Oct 01 '20 at 16:53
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    This question seems fine for me, I don't know why it was downvotes. I demand downvoter to comeback and write its reason for downvote. also I demand downvoter to pay attention to this question and remove their downvote if an edit is made – L F Oct 01 '20 at 17:07

1 Answers1

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We have that by $k=n-i$

$$\sum_{i=1}^{n} \frac{(n-i)(n-i+1)}{2}=\sum_{k=n-1}^{0} \frac{k(k+1)}{2}=\sum_{k=0}^{n-1} \frac{k(k+1)}{2}$$

user
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