Dùng phương pháp quy nạp chứng minh rằng Sn =1+5+9+...+4n-3 ( 2n-1)

$$\eqalign{ & CMR:\,\,{S_n} = 1 + 5 + 9 + ... + 4n - 3 = n\left( {2n - 1} \right)\,\,\forall n \ge 1;\,\,n \in N \cr & Voi\,\,n = 1:\,\,{S_1} = 1 = 1\left( {2 - 1} \right)\,\,\,\left( {Dung} \right). \cr & Gia\,\,su\,\,dang\,\,thuc\,\,dung\,\,voi\,\,n = k \cr & \Rightarrow {S_k} = 1 + 5 + ... + 4k - 3 = k\left( {2k - 1} \right) \cr & Ta\,\,CM\,\,dang\,\,thuc\,\,dung\,\,voi\,\,n = k + 1 \cr & Ta\,\,co:\,\,{S_{k + 1}} = 1 + 5 + ... + 4k - 3 + 4\left( {k + 1} \right) - 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = k\left( {2k - 1} \right) + 4\left( {k + 1} \right) - 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{k^2} - k + 4k + 4 - 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{k^2} + 3k + 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {k + 1} \right)\left( {2k + 1} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {k + 1} \right)\left[ {2\left( {k + 1} \right) - 1} \right] \cr & Vay\,\,dang\,\,thuc\,\,\,dung\,\,\,\forall n \ge 1;\,\,n \in N. \cr} $$