\(B=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}\left(\frac{1}{7}-\frac{1}{11}\right)+....+\frac{1}{4}\left(\frac{1}{107}-\frac{1}{111}\right)\)
\(=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{107}-\frac{1}{111}\right)\)
\(=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{111}\right)\)
\(=\frac{107}{444}\)
vậy biểu thức trên =\(\frac{107}{444}\)
B=\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}\)
\(B=\frac{111}{333}-\frac{3}{333}\)
\(B=\frac{108}{333}\)
\(B=\frac{12}{37}\)
Vậy B =\(\frac{12}{37}\)
