Explicație pas cu pas:
[tex]a = \sqrt{40} \\ \sqrt{36} < \sqrt{40} < \sqrt{49} \iff 6 < \sqrt{40} < 7 \\ \implies \Big[a\Big] = 6[/tex]
[tex]a_{n} = \frac{4n + 7}{2n + 1} = \frac{4n + 2 + 5}{2n + 1} = \frac{2(2n + 1) + 5}{2n + 1} = \\ = \frac{2(2n + 1)}{2n + 1} + \frac{5}{2n + 1} = 2 + \frac{5}{2n + 1}[/tex]
[tex]n = 0 \implies a_{0} = 7 \implies \Big[a_{0}\Big] = 7 \\ [/tex]
[tex]n = 1 \implies a_{1} = \frac{11}{3} = 3 \frac{2}{3} \implies \Big[a_{1}\Big] = 3 \\ [/tex]
[tex]n = 2 \implies a_{2} = 3 \implies \Big[a_{2}\Big] = 3 \\ [/tex]
[tex]n \geqslant 3 \implies 2n + 1 < 5 \implies \Big[a_{n}\Big] = 2 \\ [/tex]
