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

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

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