\(A=\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}\)
\(\frac{4}{5}.A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right).\left(4n+3\right)}\)
\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)
\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{4n+3}\)
\(\frac{4}{5}.A=\frac{4n+3}{12n+9}-\frac{3}{12n+9}\)
\(\frac{4}{5}.A=\frac{4n}{12n+9}\)
\(A=\frac{4n}{12n+9}:\frac{4}{5}\)
\(A=\frac{4n}{12n+9}.\frac{5}{4}\)
\(A=\frac{5n}{12n+9}\)
Đề bài sai nha bn
Ủng hộ mk nha ^_^