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Proof

(related to Proposition: Antidifferences of Some Functions)

Ad $(1)$

• In difference operators of some functions, we saw that $\Delta x^{\underline {n+1}}=(n+1)x^{\underline n}$.
• For $n\neq -1$, we get $x^{\underline n}=1/(n+1)\cdot \Delta x^{\underline {n+1}}.$
• Summing both sides yields with the linearity of indefinite sums to the formula $$\sum x^{\underline{n}}=\frac{x^{\underline{n+1}}}{n+1}.$$

Ad $(2)$

• Again, difference operators of some functions show that $\Delta a^x=a^x(a-1).$
• For $a\neq 1$, we get $a^x=1/(a-1)\cdot\Delta a^x.$
• Summing both sides yields with the linearity of indefinite sums to the formula $$\sum a^x=\frac{a^x}{a-1}.$$

Ad $(3)$

• Similarly, summing both sides of the result $\Delta 2^x=2^x$ yields $$\sum 2^x=2^x.$$

Ad $(4)$

• We calculate the difference operator $\Delta\cos(ax+b)=-2\sin(a/2)\sin(ax+b+a/2).$
• Set $b=-a/2.$
• It follows $\cos(ax-a/2)=-2\sin(a/2)\sin(ax).$
• Thus, for $a\neq 2n\pi$: $$\sum\sin(ax)=-\frac{\cos(a(x-1/2))}{2\sin(a/2)}.$$

Ad $(5)$

• Similarly, from $\Delta\sin(ax+b)=-2\sin(a/2)\cos(ax+b+a/2),$ we deduce for $a\neq 2n\pi$: $$\sum\cos(ax)=\frac{\sin(a(x-1/2))}{2\sin(a/2)}.$$

Ad $(6)$

• In difference operators of some functions, we saw that $\Delta(a+bx)^{\underline {n+1}}=b(n+1)$.
• For $n\neq -1$, we get $(a+bx)^\underline{n}=1/(b(n+1))\cdot \Delta(a+bx)^{\underline {n+1}}.$
• Summing both sides yields with the linearity of indefinite sums to the formula $$\sum(a+bx)^{\underline{n}}=\frac{(a+bx)^{\underline{n+1}}}{b(n+1)}.$$
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References

Bibliography

1. Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960