\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(A=\frac{3+1}{3}.\frac{8+1}{8}.\frac{15+1}{15}...\frac{n^2+2n+1}{n^2+2n}\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{\left(n+1\right)^2}{n^2+2n}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(A=\left(n+1\right).\frac{2}{n+2}=\frac{2.\left(n+1\right)}{n+2}\)
Ta có : \(1+\frac{1}{k^2+2k}=\frac{k^2+2k+1}{k^2+2k}=\frac{\left(k+1\right)^2}{k\left(k+2\right)}\) với k thuộc N*
Áp dụng với k = 1,2,3,....,n được :
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{\left(1+1\right)^2}{1.\left(1+2\right)}.\frac{\left(2+1\right)^2}{2.\left(2+2\right)}.\frac{\left(3+1\right)^2}{3.\left(3+2\right)}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(=\frac{\left[2.3.4...\left(n+1\right)\right]^2}{1.2.3...n.3.4.5...\left(n+2\right)}=\frac{\left[\left(n+1\right)!\right]^2}{n!.\frac{\left(n+2\right)!}{2}}\)
