Giải bài 1 trang 92 – SGK môn Đại số và Giải tích lớp 11

Gợi ý: 

Thay $n=1, 2, 3, 4, 5$ vào số hạng tổng quát rồi tính.

a) $u_1=1;\,\,\,u_2=\dfrac{2}{3};\,\,\,u_3=\dfrac{3}{7};\,\,\,u_4=\dfrac{4}{15};\,\,\,u_5=\dfrac{5}{31}$

b) ${{u}_{1}}=\dfrac{1}{3};\,\,\,{{u}_{2}}=\dfrac{3}{5};\,\,\,{{u}_{3}}=\dfrac{7}{9};\,\,\,{{u}_{4}}=\dfrac{15}{17};\,\,\,{{u}_{5}}=\dfrac{31}{33}$

c) ${{u}_{1}}=2;\,\,\,{{u}_{2}}=\dfrac{9}{4};\,\,\,{{u}_{3}}={{\left( \dfrac{4}{3} \right)}^{3}}=\dfrac{64}{27};\,\,\,{{u}_{4}}={{\left( \dfrac{5}{4} \right)}^{4}}=\dfrac{625}{256};\,\,\,{{u}_{5}}={{\left( \dfrac{6}{5} \right)}^{5}}=\dfrac{7776}{3125}$

d) ${{u}_{1}}=\dfrac{1}{\sqrt{2}};\,\,\,{{u}_{2}}=\dfrac{2}{\sqrt{5}};\,\,\,{{u}_{3}}=\dfrac{3}{\sqrt{10}};\,\,\,{{u}_{4}}=\dfrac{4}{\sqrt{17}};\,\,\,{{u}_{5}}=\dfrac{5}{\sqrt{26}}$