[-[S]ycoraX-]

ข้อความเดิมเขียนโดยคุณ [G]enerate

จงหาค่าของ $( \frac{3}{4} + \frac{5}{36}+ \frac{7}{144} + \frac{9}{400} + \frac{11}{900} + \frac{13}{1764} )42^2 $

$( \frac{3}{4} + \frac{5}{36}+ \frac{7}{144} + \frac{9}{400} + \frac{11}{900} + \frac{13}{1764} )42^2 $ = $( [\frac{1}{1^2}-\frac{1}{2^2}] +[ \frac{1}{2^2}-\frac{1}{3^2}]+ [\frac{1}{3^2}-\frac{1}{4^2}] + [\frac{1}{4^2}-\frac{1}{5^2}] + [\frac{1}{5^2}-\frac{1}{6^2}] +[\frac{1}{6^2}-\frac{1}{7^2}] )42^2 $

=($\frac{1}{1^2}-\frac{1}{7^2})42^2$
=$12^3$
=1728 หรือเปล่าครับ