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I'm doing a homework assignment which requires us to merge multiple summations into a single sum before computing it. The given problem is: $$\sum_{i=1}^{100} i^2 + \sum_{i=1}^{101} i + \sum_{i=k}^{102} k + \sum_{i=1}^{103} 1$$

I know the following properties of summation where:$$\sum_{k=m}^{n} a_k = \sum_{k=m}^{n-1} a_k + a_n$$ and $$\sum_{k=m}^{n} a_k + \sum_{k=m}^{n} b_k = \sum_{k=m}^{n} (a_k +b_k)$$

but I'm still having trouble using them to get a single summation. Help would be much appreciated.

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    $\begingroup$ Did you really mean $\sum_{i=k}^{102}k$? That is $k\times (103-k)$, hence a function of $k$. $\endgroup$
    – lulu
    Oct 11, 2019 at 17:36
  • $\begingroup$ Yes, that's what the problem states and it is part of why I'm struggling to solve it $\endgroup$
    – Lin
    Oct 11, 2019 at 17:40
  • $\begingroup$ Well, each sum is easily solved on its own. Combining them in a single sum won't really work as the number of terms in each sum is different. $\endgroup$
    – lulu
    Oct 11, 2019 at 17:42
  • $\begingroup$ If the problem stated that the $\sum_{i=k}^{102} k = k+ k+ k+ k+ .... + k = m*k$ where $m = 103- k$. So that sum is $k(103-k)$. $\endgroup$
    – fleablood
    Oct 11, 2019 at 18:42
  • $\begingroup$ @lulu Can’t we write $\displaystyle\sum_{i=1}^{100} i^2 + \sum_{i=1}^{101} i + \sum_{i=k}^{102} k + \sum_{i=1}^{103} 1=\sum_{n=1}^{407-k}a_n=\underbrace{1^2+\dots+100^2+1+\dots+101+\underbrace{k+\dots +k}_{103-k\text{ times}}+\underbrace{1+\dots+1}_{103 \text{ times}}}_{n=1\text{ to }407-k}$ since the sum has $407-k$ total terms? This comment is for a related problem. $\endgroup$ Sep 30, 2022 at 12:48

1 Answer 1

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I don't know if this is meant to be solved in this manner or not, but I don't think you can solve this in a conventional way. But you can also do this out of the way;

$$\sum_{i=1}^{103}\left(i^2\left(1-\left \lfloor{\frac{i}{101}}\right \rfloor\right) + i\left(1-\left \lfloor{\frac{i}{102}}\right \rfloor\right) +(k+i)\left(1-\left \lfloor{\frac{i}{102}}\right \rfloor\right)+ 1\right)$$

Good luck trying to compute from this.

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