\(A=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}+...+\frac{1}{n^2}-\frac{1}{4n}=\frac{56}{673}\)
\(\Rightarrow4A=\)
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n\left(n+4\right)}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{56}{673}:\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{3}-\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{2019}\)
=> n + 4 = 2019
n = 2019 - 4
n = 2015
