a. Ta có
\(\eqalign{
& {u_1} = {{{{2.1}^2} - 3} \over 1} = - 1 \cr
& {u_2} = {{{{2.2}^2} - 3} \over 2} = {5 \over 2} \cr
& {u_3} = {{{{2.3}^2} - 3} \over 3} = 5 \cr
& {u_4} = {{{{2.4}^2} - 3} \over 4} = {{29} \over 4} \cr
& {u_5} = {{{{2.5}^2} - 3} \over 5} = {{47} \over 5} \cr} \)
b.
\(\eqalign{
& {u_1} = {\sin ^2}{\pi \over 4} + \cos {{2\pi } \over 3} = {1 \over 2} - {1 \over 2} = 0 \cr
& {u_2} = {\sin ^2}{\pi \over 2} + \cos {{4\pi } \over 3} = 1 - {1 \over 2} = {1 \over 2} \cr
& {u_3} = {\sin ^2}{{3\pi } \over 4} + \cos 2\pi = {1 \over 2} + 1 = {3 \over 2} \cr
& {u_4} = {\sin ^2}\pi + \cos {{8\pi } \over 3} = \cos \left( {2\pi + {{2\pi } \over 3}} \right) = - {1 \over 2} \cr
& {u_5} = {\sin ^2}{{5\pi } \over 4} + \cos {{10\pi } \over 3} = {1 \over 2} - {1 \over 2} = 0 \cr} \)
c.
\(\eqalign{
& {u_1} = - 2 \cr
& {u_2} = 4 \cr
& {u_3} = - 8 \cr
& {u_4} = 16 \cr
& {u_5} = - 32 \cr} \)
