Răspuns:
[tex]\it a)~~\boxed{\boxed{\dfrac{7}{8};~\dfrac{9}{10};~\dfrac{11}{12};~\dfrac{13}{14};.....}}[/tex]
[tex]\it b)~~\boxed{\boxed{\it T_{50} =\dfrac{99}{100}}}[/tex]
[tex]\it c)~~ \boxed{\boxed{\it T_{n-1} =\dfrac{2n-3}{2n-2}}}[/tex]
Explicație pas cu pas:
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[tex]\it~~~~~~~~~~~[/tex]
[tex]\it \dfrac{1}{2};~\dfrac{3}{4};~ \dfrac{5}{6}; .......................[/tex]
[tex]\it~~~~~~~~~[/tex]
[tex]\it T_{1} =\dfrac{1}{2}~\xrightarrow[a~termenului]{alta~scriere}~T_{1} = \dfrac{2\cdot 1 - 1}{2\cdot 1}[/tex]
[tex]\it T_{2} =\dfrac{3}{4}~\xrightarrow[a~termenului]{alta~scriere}~T_{2} = \dfrac{ 2\cdot2-1}{2\cdot2}[/tex]
[tex]\it T_{3} =\dfrac{5}{6}~\xrightarrow[a~termenului]{alta~scriere}~T_{3} = \dfrac{ 2\cdot3-1}{2\cdot3}[/tex]
[tex]\it T_{4} =\dfrac{7}{8}~\xrightarrow[a~termenului]{alta~scriere}~T_{4} = \dfrac{ 2\cdot4-1}{2\cdot4}[/tex]
[tex]\it T_{5} =\dfrac{9}{10}~\xrightarrow[a~termenului]{alta~scriere}~T_{5} = \dfrac{ 2\cdot5-1}{2\cdot5}[/tex]
[tex]\it T_{6} =\dfrac{11}{12}~\xrightarrow[a~termenului]{alta~scriere}~T_{6} = \dfrac{ 2\cdot6-1}{2\cdot 6}[/tex]
[tex]\it T_{7} =\dfrac{13}{14}~\xrightarrow[a~termenului]{alta~scriere}~T_{7} = \dfrac{ 2\cdot7-1}{2\cdot 7}[/tex]
[tex]\it T_{8} =\dfrac{15}{16}~\xrightarrow[a~termenului]{alta~scriere}~T_{8} = \dfrac{ 2\cdot8-1}{2\cdot 8}[/tex]
[tex]\it..............................................[/tex]
[tex]\boxed{\boxed{\it T_{n} = \dfrac{2\cdot n-1}{2\cdot n}\rightarrow forma~termenului~general}}[/tex]
[tex]\it T_{50} = \dfrac{2\cdot 50-1}{2\cdot 50}\implies T_{50} = \dfrac{100-1}{100}\implies \boxed{\boxed{\it T_{50} =\dfrac{99}{100}}}[/tex]
[tex]\it T_{n-1} = \dfrac{2\cdot (n-1)-1}{2\cdot (n-1)}\implies T_{n-1} = \dfrac{2n-2-1}{2n-2}\implies \boxed{\boxed{\it T_{n-1} =\dfrac{2n-3}{2n-2}}}[/tex]
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