[tex]a_{n} =\dfrac{4n-5}{2n+3} ~~\land~~a_{n} =1~~\Rightarrow ~~\dfrac{4n-5}{2n+3}=1\\\\\dfrac{4n-5}{2n+3}=1\\\\4n-5=2n+3\\\\4n-2n=3+5\\\\2n=8~~\mid \div 2\\\\n=4~~\land~~n\in N_{+} ~~\Rightarrow ~~n=4\\\\Sprawdzam:\\\\a_{4} =\dfrac{4\cdot 4-5}{2\cdot 4+3}\\\\a_{4} =\dfrac{16-5}{8+3}\\\\a_{4} =\dfrac{11}{11}\\\\a_{4} =1\\[/tex]
Odp: Czwarty wyraz ciągu jest równy 1.
Poprawna odopwiedź: C. 4
