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The summation goes as follows: 0 + 1 + 3 + 6 + 10 + 15 ... and $k $ starts at 1 where k is the current iteration and where the next number is equal to the previous number plus the previous iteration $k$.

KetDog
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  • And what have you tried? – JBL Oct 29 '22 at 02:10
  • @JBL really don't know how would I come up with a closed form for a summation like this since I don't have strong mathematical background, but I would be willing to try if I knew where to start. This sequence I saw while doing some programming assignment and I got curious. – KetDog Oct 29 '22 at 02:11
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    This series is interesting. If you do $0 + 1$, $3 + 6$, $10 + 15$, $21+28$ you see it forms a series of perfect squares. Except that it omits all even perfect squares. So what you really have is $1 + 9 + 25 + 49 + 81 + \cdots k$. – Bumblebee Oct 29 '22 at 02:16
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    Equivalently (and omitting the zero term) you can group it as $1+3$, $6+10$, $15+21$, ... to get all the even perfect squares. – catherine Oct 29 '22 at 02:18
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    The sum of $n$ consecutive squares is: $$1^2 + 2^2 + 3^2 + 4^2 + \cdots n^2 = \frac{n(n+1)(2n+1)}{6}.$$ Consider how grouping the series as odd and even perfect squares can "fix" the formula. – Bumblebee Oct 29 '22 at 02:29
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    If we take the sequence of the differences $a_n-a_{n-1}$ and repeat this process and eventually we arrive at a constant sequence (this is the case here) , then there is a polynomial completely describing this sequence. The polynomial can easily be found , for example , by interpolation. – Peter Oct 29 '22 at 08:34

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