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If i have a sum say

$$C\sum_{l=1}^C\left( a+b\cdot c_l\right)$$

Is is true in general that $$C\sum_{l=1}^C\left( a+b\cdot c_l\right) = a+\left( C\sum_{l=1}^C\left(b\cdot c_l \right) \right)$$ and $$C\sum_{l=1}^C\left( a+b\cdot c_l\right) = a+b\left( C\sum_{l=1}^C\left(c_l\right) \right)$$

Sorry if this has been asked before i wasn't able to find this specific identity.

EDIT: $$\frac{1}{C}\sum_{l=1}^C\left( a+b\cdot c_l\right) = a+\left( \frac{1}{C}\sum_{l=1}^C\left(b\cdot c_l \right) \right)\label{1}$$

StubbornAtom
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1 Answers1

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What you wrote is true if you replace your first terms of $a$ with $C^2(a)$. This is because you can split your summation into $2$ parts, i.e.,

$$C\sum_{l=1}^C\left( a+b\cdot c_l\right) = C\sum_{l=1}^C a + C\sum_{l=1}^C\left(b\cdot c_l\right) \tag{1}\label{eq1A}$$

With the first term on the right side, since you are summing a constant of $a$ a total of $C$ times, you have

$$C\sum_{l=1}^C a = C(Ca) = (C^2)a \tag{2}\label{eq2A}$$

John Omielan
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  • Thank you. So if it would have been $\frac{1}{C} \sum_{l=1}^C(a+b+c_l)$ it would have been true, right? –  Mar 01 '20 at 01:49
  • @sn3jd3r You're welcome. Yes, with your new equation, the first terms would then be correct, i.e., being $a$, but you'd of course need to adjust the second terms accordingly. – John Omielan Mar 01 '20 at 01:51
  • Okay - just to be sure by adjusting the second term, you mean like it is done in the EDIT: of the question? @john-omielan –  Mar 01 '20 at 01:55
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    @sn3jd3r Yes, that is absolutely correct. – John Omielan Mar 01 '20 at 01:55
  • Since summation is on $l$ so, $C\sum_{l=1}^C a = Ca$ how it can be $C^{2}a$ ? – Akash Patalwanshi Dec 01 '23 at 13:34
  • @AkashPatalwanshi Although the summation value of $l$ isn't used in the expression being summed, the summation is still going from $1$ to $C$, i.e., there are $C$ terms being summed, with each term being just $a$, so the sum is $Ca$. Another way to consider this is the summation expression, i.e., $a$, is a function of $l$, e.g., the constant function $f(l)=a$ or, to have $l$ being used, $f(l)=a(l^{0})$, in both cases for all $1\le l\le C$. We then get $\sum_{l=1}^{C}f(l) = f(1) + f(2) + \ldots + f(C) = a + a + \ldots + a = Ca$. Thus, $C\sum_{l=1}^{C}a = C\sum_{l=1}^{C}f(l) = C(Ca) = C^{2}a$. – John Omielan Dec 01 '23 at 16:21