$\begin{array}{l}
(1 - \dfrac{1}{{{2^2}}})(1 - \dfrac{1}{{{3^2}}})(1 - \dfrac{1}{{{4^2}}}).....(1 - \dfrac{1}{{{n^2}}}) = \dfrac{{1024}}{{2047}}\\
 <  =  > \dfrac{{(4 - 1)(9 - 1)(16 - 1)....(n - 1)(n + 1)}}{{{2^2}{{.3}^2}{{.4}^2}.....{n^2}}} = \dfrac{{1024}}{{2047}}\\
 <  =  > \dfrac{{1.3.2.4.3.5.......n(n + 2)(n - 1)(n + 1)}}{{{2^2}{{.3}^2}{{.4}^2}.....{{(n - 1)}^2}{n^2}}} = \dfrac{{1024}}{{2047}}\\
 <  =  > \dfrac{{n + 1}}{{2n}} = \dfrac{{1024}}{{2047}}\\
 <  =  > \dfrac{1}{2} + \dfrac{1}{{2n}} = \dfrac{{1024}}{{2047}} =  > n = 2047
\end{array}$

                                                           mik k biết bạn ak

bạn viết cách giải bằng typemath được không ?

(1-1/2^2).....(1-1/n^2)=(2^2-1)/2^2.............(n^2-1)/n^2=1*3/2^2........(n-1)(n+1)/n^2=(n+1)/2n=>2047