Folosind formula lui Gauss, vom avea:
[tex]\it S=\dfrac{1}{\dfrac{2\cdot3}{2}}+\dfrac{1}{\dfrac{3\cdot4}{2}}+\dfrac{1}{\dfrac{4\cdot5}{2}}+\ ...\ +\dfrac{1}{\dfrac{n\cdot(n+1)}{2}}=\\ \\ \\ =\underbrace{\it \dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+ \ ...\ +\dfrac{2}{n\cdot(n +1)}}_{n-1\ termeni}[/tex]
Așadar, numărul de termeni căutat este n - 1 .
[tex]\it S > 0,9 \Rightarrow S > \dfrac{9}{10} \Rightarrow 2\Big( \dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\ ...\ +\dfrac{1}{n\cdot(n+1)}\Big) > \dfrac{9}{10} \Rightarrow\\ \\ \\ \Rightarrow2\Big(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\ ...\ +\dfrac{1}{n}-\dfrac{1}{n+1}\Big) > \dfrac{9}{10} \Rightarrow[/tex]
[tex]\it \Rightarrow2\Big(\dfrac{1}{2}-\dfrac{1}{n+1}\Big) > \dfrac{9}{10} \Rightarrow2\cdot\dfrac{n+1-2}{2(n+1)} > \dfrac{9}{10} \Rightarrow \dfrac{n-1}{n+1} > \dfrac{9}{10} \Rightarrow \\ \\ \\ \Rightarrow10n-10 > 9n+9 \Rightarrow 10n-9n > 9+10 \Rightarrow n > 19\Big|_{-1} \Rightarrow n-1 > 18 \Rightarrow \\ \\ \\ \Rightarrow n-1\geq19[/tex]
Cel mai mic număr de termeni pentru care S > 0,9 este egal cu 19 .
