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Proof

(related to Proposition: Basic Calculations Involving Indefinite Sums)

Ad $(1)$

• Let $f$ be a complex-valued function $f:\mathbb C\to C$ and let $\lambda\in\mathbb C$ be a constant.
• Choose an antidifference $F:\mathbb C\to\mathbb C$, that is $\Delta F(x)=f(x)$.
• It follows $\lambda\cdot \Delta F(x)=c\cdot f(x)=\Delta (cF)(x),$ applying basic calculations involving the difference operator (no. 1).
• Summing on both sides yields $\sum cf(x)=cF(x)=c\sum f(x).$

Ad $(2)$

• Chose the antiderivatives $F,G:\mathbb C\to\mathbb C$ of $f,g$, i.e. $\Delta F(x)=f(x)$ and $\Delta G(x)=g(x)$.
• Applying basic calculations involving the difference operator (no. 2), we get $\Delta (F(x)\pm G(x))=\Delta F(x)\pm \Delta G(x).$
• Summing on both sides yields $(F(x)\pm G(x))=F(x)\pm G(x).$
• This means $\sum (f(x)\pm g(x))=\sum f(x)\pm \sum g(x).$

Ad $(3)$

• The product rule of the difference operator, can be re-written for suitable complex-valued functions $f,g$ as follows: $g(x)\Delta f(x)=\Delta (fg)(x)-f(x+1)\Delta g(x).$
• Summing on both sides yields $\sum g(x)\Delta f(x)=f(x)g(x)-\sum f(x+1)\Delta g(x).$
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