a) \(S_1=\dfrac{1}{1.5}=\dfrac{1}{5}\) \(S_2=\dfrac{1}{1.5}+\dfrac{1}{5.9}=\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{5}\right)+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}\right)\) \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}\right)=\dfrac{1}{4}\left(1-\dfrac{1}{9}\right)=\dfrac{2}{9}\). \(S_3=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}\right)\) \(=\dfrac{1}{4}\left(1-\dfrac{1}{13}\right)=\dfrac{3}{13}\). \(S_4=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}\)\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\) \(=\dfrac{1}{4}\left(1-\dfrac{1}{17}\right)=\dfrac{4}{17}\). b) Dự đoán công thức : \(S_n=\dfrac{1}{4}\left(1-\dfrac{1}{4n+1}\right)\). Chứng minh bằng quay nạp: Với \(n=1\): \(S_1=\dfrac{1}{1.5}=\dfrac{1}{5}\). Vậy giả thiết quy nạp đúng với n = 1. Giả sử điều cần chứng minh đúng với \(n=k\). Nghĩa là: \(S_k=\dfrac{1}{4}\left(1-\dfrac{1}{4k+1}\right)\). Ta sẽ chứng minh nó đúng với \(n=k+1\): \(S_{k+1}=\dfrac{1}{4}\left(1-\dfrac{1}{4\left(k+1\right)+1}\right)\) Thật vậy: \(S_{k+1}=S_k+\dfrac{1}{\left[4\left(k+1\right)-3\right].\left[4\left(k+1\right)+1\right]}\) \(=\dfrac{1}{4}\left(1-\dfrac{1}{4k+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{4\left(k+1\right)-3}-\dfrac{1}{4\left(k+1\right)+1}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{4k+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{4k+1}-\dfrac{1}{4\left(k+1\right)+1}\right)\) \(=\dfrac{1}{4}\left(1-\dfrac{1}{4\left(k+1\right)+1}\right)\). Vậy điều cần chứng minh đúng với mọi n.