[tex]\displaystyle\bf\\\\Folosim~formula:\\\\\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\\\\\S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+~\cdots~+\frac{1}{97\cdot98}+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\\\\\\S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\\\\Reducem~termenii:\\\\S=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{100-1}{100}=\boxed{\bf\frac{99}{100}<1}[/tex]
⇒ S este subunitar.
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