Clarity on the following summation rule

Question

Consider the following summation rule:

$$\sum_{k=1}^n c=nc$$ given that this is true, it implies that $$\sum_{k=0}^{n}c=(n+1)c$$ and the question is, does this imply that the general case is $$\sum_{k=r}^nc=(n-r+1)c$$ I'm asking because I had to compute $$\sum_{k=20}^{100}(x+n)$$ and found that the answer was $81x+4860$ instead of $80x+4860$ as I expected it would be, and hence I made the inference that what I stated above is true. Is it true for all cases of $r,k$ and $c$?

Thank you.

$\sum_{k=1}^n c_n = nc_n$ or $\sum_{k=1}^n c = nc$ – Randall Nov 03 '17 at 14:22 — Randall, Nov 03 '17 at 14:22
yes if $ n \ge r$ otherwise it is debatable. – amcalde Nov 03 '17 at 14:26 — amcalde, Nov 03 '17 at 14:26

score 1 · Accepted Answer · answered Nov 03 '17 at 14:37

I think your second equation$\sum_{r=k}^n c = (r-k+1)c \tag{1}$ should be $\sum_{r=k}^n c = (n-k+1)c$. Your last equation $\sum_{k=20}^{100}(x+n)$ should be $\sum_{k=20}^{100}(x+k) \tag{2}$. The result is $(100-20+1)(x) + \sum_{k=20}^{100} k $ = 81x+4860. So your general summation is right.

Clarity on the following summation rule

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