Răspuns :
[tex]\displaystyle\bf\\1)\\1+2+2^2+2^3+...+2^{2020}=\\=2^0+2^1+2^2+2^3+...+2^{2020}=\frac{2^{2020+1}}{2-1}=\boxed{\bf2^{2021}}\\\\[/tex]
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[tex]\displaystyle\bf\\2)\\\\\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+...+\frac{1}{\sqrt{9}+\sqrt{8}}=\\\\~~~(Propun~sa~le~scriu~pe~toate~ca~sunt~putine)\\\\=\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}+\\\\+\frac{1}{\sqrt{6}+\sqrt{5}}+\frac{1}{\sqrt{7}+\sqrt{6}}+\frac{1}{\sqrt{8}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{8}}=\\\\Rationalizam~numitorii.[/tex]
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[tex]\displaystyle\bf\\=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+\frac{\sqrt{5}-\sqrt{4}}{5-4}+\\\\+\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{\sqrt{7}-\sqrt{6}}{7-6}+\frac{\sqrt{8}-\sqrt{7}}{8-7}+\frac{\sqrt{9}-\sqrt{8}}{9-8}=\\\\\\=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\\\\+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}=\\\\Se~reduc~doua~cate~doua~si~ramine:\\\\=-\sqrt{1}+\sqrt{9}=-1+3=\boxed{\bf2}[/tex]