We know the sum-series formula, n(n + 1) / 2, to get the sum of the natural numbers less than n. This means that we can put an upper limit of n^2 on the result with respect to n. If we want the sum of the first m elements of this series (i.e. https://www.geeksforgeeks.org/sum-of-sum-series-of-first-n-natural-numbers/), can we put an upper limit on that result with respect to m? How would one do this?

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    $\begingroup$ Unless I'm misunderstanding your question, the article you linked gives a formula for the sum of a sum-series. $\endgroup$ May 15, 2021 at 3:03
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    $\begingroup$ There is an expression for the sum of the first $n$ squares which would actually help you to compute $\sum_{n=1}^m \frac{n(n+1)}{2}$ exactly. It can also help if you want to compute the upper bound $\sum_{n=1}^m n^2$ that you mention. $\endgroup$
    – angryavian
    May 15, 2021 at 3:04

1 Answer 1


For the sum of sum-series, we need,

$$\begin{align}\sum_{n=1}^m\frac{n(n+1)}{2} &= \frac12\sum_{n=1}^m(n^2+n) \\ &= \frac12\left[\frac{m(m+1)(2m+1)}{6}+\frac{m(m+1)}2 \right]\\ &=\frac{m(m+1)(m+2)}{6}\end{align}$$

Could you fix the bounds now?


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