a) Kiểm tra với \(n = 1,\) ta có \({A_1} = \dfrac{1}{{1.2.3}} = \dfrac{1}{6} = \dfrac{{1.\left( {1 + 3} \right)}}{{4.2.3}}\).
Giả sử ta có \({A_k} = \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + ... + \dfrac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}} = \dfrac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}\)
Ta cần chứng minh \({A_{k + 1}} = \dfrac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}\)
Thật vậy,
\({A_{k + 1}} = {A_k} + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\) \( = \dfrac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\)
\( = \dfrac{{k{{\left( {k + 3} \right)}^2} + 4}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\) \( = \dfrac{{{k^3} + 6{k^2} + 9k + 4}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\) \( = \dfrac{{\left( {k + 4} \right){{\left( {k + 1} \right)}^2}}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\) \( = \dfrac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}\)
Vậy ta có điều phải chứng minh.
b) Kiểm tra với \(n = 1\) ta có \({B_1} = \dfrac{{1.\left( {1 + 1} \right)}}{2} = 1 = \dfrac{{1\left( {1 + 1} \right)\left( {1 + 2} \right)}}{6}\) nên \(n = 1\) đúng.
Giả sử đã có \({B_k} = \dfrac{{k\left( {k + 1} \right)\left( {k + 2} \right)}}{6}.\)
Ta cần chứng minh \({B_{k + 1}} = \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{6}\)
Thật vậy,
\({B_{k + 1}} = {B_k} + \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2}\) \( = \dfrac{{k\left( {k + 1} \right)\left( {k + 2} \right)}}{6} + \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2}\) \( = \dfrac{{k\left( {k + 1} \right)\left( {k + 2} \right) + 3\left( {k + 1} \right)\left( {k + 2} \right)}}{6}\) \( = \dfrac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{6}\)
Vậy ta có điều phải chứng minh.
c) Kiểm tra với \(n = 1\) ta có: \({S_1} = \sin x = \dfrac{{\sin \dfrac{x}{2}.\sin x}}{{\sin \dfrac{x}{2}}}\) nên đúng.
Giả sử đã có \({S_k} = \dfrac{{\sin \dfrac{{kx}}{2}.\sin \dfrac{{\left( {k + 1} \right)}}{2}x}}{{\sin \dfrac{x}{2}}}.\)
Ta cần chứng minh \({S_{k + 1}} = \dfrac{{\sin \dfrac{{\left( {k + 1} \right)x}}{2}.\sin \dfrac{{\left( {k + 2} \right)x}}{2}}}{{\sin \dfrac{x}{2}}}\)
Thật vậy,
\({S_{k + 1}} = {S_k} + \sin \left( {k + 1} \right)x\) \( = \dfrac{{\sin \dfrac{{kx}}{2}.\sin \dfrac{{\left( {k + 1} \right)}}{2}x}}{{\sin \dfrac{x}{2}}} + \sin \left( {k + 1} \right)x\) \( = \dfrac{{ - \dfrac{1}{2}\left( {\cos \dfrac{{\left( {2k + 1} \right)x}}{2} - \cos \dfrac{x}{2}} \right) - \dfrac{1}{2}\left( {\cos \dfrac{{\left( {2k + 3} \right)x}}{2} - \cos \dfrac{{\left( {2k + 1} \right)x}}{2}} \right)}}{{\sin \dfrac{x}{2}}}\)
\( = \dfrac{{ - \dfrac{1}{2}\left( {\cos \dfrac{{\left( {2k + 3} \right)x}}{2} - \cos \dfrac{x}{2}} \right)}}{{\sin \dfrac{x}{2}}}\) \( = \dfrac{{\sin \dfrac{{\left( {k + 2} \right)x}}{2}\sin \dfrac{{\left( {k + 1} \right)x}}{2}}}{{\sin \dfrac{x}{2}}}\)
Vậy \({S_{k + 1}} = \dfrac{{\sin \dfrac{{\left( {k + 1} \right)x}}{2}.\sin \dfrac{{\left( {k + 2} \right)x}}{2}}}{{\sin \dfrac{x}{2}}}\left( {dpcm} \right).\)
