$S_n =\dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} +\dots +\dfrac{1}{(2n-1)(2n+1)}$

a) $S_1=\dfrac{1}{1.3} =\dfrac13$

$S_2=\dfrac{1}{1.3} +\dfrac{1}{3.5} =\dfrac13 +\dfrac{1}{15} = \dfrac25$

$S_3 =\dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} = \dfrac37$

$S_4 =\dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} +\dfrac{1}{7.9} = \dfrac49$

b) Dự đoán $S_n$

$S_n =\dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} +\dots +\dfrac{1}{(2n-1)(2n+1)}$

$\to 2S_n =\dfrac{2}{1.3} +\dfrac{2}{3.5} +\dfrac{2}{5.7} +\dots +\dfrac{2}{(2n-1)(2n+1)}$

$\to 2S_n =\dfrac{3 -1}{1.3} +\dfrac{5-3}{3.5} +\dfrac{7-5}{5.7} +\dots +\dfrac{(2n +1) - (2n -1)}{(2n-1)(2n+1)}$

$\to 2S_n =\dfrac{3}{1.3} -\dfrac{1}{1.3}+\dfrac{5}{3.5} -\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7} +\dots +\dfrac{2n+1}{(2n-1)(2n+1)}-\dfrac{2n-1}{(2n-1)(2n+1)}$

$\to 2S_n =1 -\dfrac{1}{3}+\dfrac{1}{3} -\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7} +\dots +\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$

$\to 2S_n =1 -\dfrac{1}{2n +1}$

$\to S_n =\dfrac12 -\dfrac{1}{2(2n+1)}$

$\to S_n = \dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} +\dots +\dfrac{1}{(2n-1)(2n+1)} = \dfrac12 -\dfrac{1}{2(2n+1)}\qquad (*)$

Chứng minh:

+) Với $n = 1$ ta được:

$\dfrac13 = \dfrac13$ (đúng)

+) Giả sử $(*)$ đúng với $n = k \geq 1$ tức là:

$S_k = \dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} +\dots +\dfrac{1}{(2k-1)(2k+1)} = \dfrac12 -\dfrac{1}{2(2k+1)}$ (giả thiết quy nạp)

+) Cần chứng minh $(*)$ đúng với $n = k +1$, tức là cần chứng minh:

$S_{k+1} = \dfrac{1}{1.3} +\dfrac{1}{3.5} +\dfrac{1}{5.7} +\dots +\dfrac{1}{(2k-1)(2k+1)} +\dfrac{1}{[2(k+1) -1][2(k+1) +1]}= \dfrac12 -\dfrac{1}{2[2(k+1)+1]}$

Thật vậy:

$S_{k+1} = S_k + \dfrac{1}{[2(k+1) -1][2(k+1) +1]} = \dfrac12 -\dfrac{1}{2(2k+1)} +\dfrac{1}{(2k +1)(2k+3)}$

$\to S_{k+1} =\dfrac12 -\dfrac{2k + 3 - 2}{2(2k +1)(2k+3)} =\dfrac12-\dfrac{1}{2(2k+3)}$

$\to S_{k+1} =\dfrac12 -\dfrac{1}{2[2(k+1) +1]}$

Vậy $S_n = \dfrac12 -\dfrac{1}{2(2n+1)} \forall n \in \Bbb N^*$