[tex]\text{Pentru prima suma se poate folosi suma lui Gauss:}\\
\boxed{1+2+3+_{\dots}+n=\dfrac{n\cdot (n+1)}{2}}\\
Asadar:1+2+3+_{\dots}+100=\dfrac{100\cdot 101}{2}=50\cdot 101=...\\
\\
1\cdot 2\cdot 3\cdot 4\cdot _{\dots}\cdot 100=100! (\ se\ citeste\ "100\ factorial")\\
\text{Nu se poate calcula pentru ca e un numar mult prea mare.}\\
\\
S=2+2^2+_{\dots}+2^{100}|\cdot 2\\
2S=2^2+2^3+_{\dots}+2^{101}\\
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2S-S=2^{101}-2\Rightarrow \boxed{S=2^{101}-2}[/tex]
[tex]Observatie:\text{Se pot folosi formulele de la progresii geometrice.}\\
S=b_1\cdot \dfrac{q^n-1}{q-1}\\
unde\ b_1\ e\ primul\ termen,q=ratia\ si\ n=numarul\ de\ termeni.[/tex]
