\(B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{1}{5}B=\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)
\(B-\frac{1}{5}B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}-\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\)\(\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+\frac{4}{11}-\frac{4}{15}+...+\frac{4}{4n-1}-\frac{4}{4n+3}\)
\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{4n-3}\)
\(\frac{4}{5}B=\frac{16n-24}{12n-9}\)
\(B=\frac{\frac{16n-24}{12n-9}}{\frac{4}{5}}\)
\(B=\frac{20n-30}{12n-9}\)
B = \(\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}+\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}.\left(\frac{1}{3}-\frac{1}{4n+3}\right)=\frac{5}{12}-\frac{5}{4\left(4n+3\right)}=\frac{5}{12}-\frac{5}{16n+12}\)
sửa lại
\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)
\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{4n+3}\)
\(\frac{4}{5}B=\frac{4n}{12n+9}\)
\(B=\frac{\frac{4n}{12n+9}}{\frac{4}{5}}\)
\(B=\frac{5n}{12n+9}\)
