22+52+82+...+(3k−1)2=k(6k2+3k−1)222+52+82+...+(3k−1)2=k(6k2+3k−1)2
Ta cần phải chứng minh (1) đúng với n=k+1
⇔22+52+82+...+(3k−1)2+(3k+2)2=(k+1)(6k2+15k+8)2⇔22+52+82+...+(3k−1)2+(3k+2)2=(k+1)(6k2+15k+8)2
=6k3+21k2+23k+82=6k3+15k2+8k+6k2+15k+82=6k3+21k2+23k+82=6k3+15k2+8k+6k2+15k+82
2+52+82+...+(3n−1)2=n(6n2+3n−1)2(1)22+52+82+...+(3n−1)2=n(6n2+3n−1)2(1)
Với n=1
VT=4;VP=4VT=4;VP=4
(1) đúng với n=1
Giả sử (1) đúng với n=k≥1k≥1
⇔22+52+82+...+(3k−1)2+[3(k+1)−1]2=(k+1)[6(k+1)2+3(k+1)−1]2⇔22+52+82+...+(3k−1)2+[3(k+1)−1]2=(k+1)[6(k+1)2+3(k+1)−1]2
VT=k(6k2+3k−1)2+(3k+2)2=6k3+3k2−k+18k2+24k+82VT=k(6k2+3k−1)2+(3k+2)2=6k3+3k2−k+18k2+24k+82
=k(6k2+15k+8)+(6k2+15k+8)2=(6k2+15k+8)(k+1)2=k(6k2+15k+8)+(6k2+15k+8)2=(6k2+15k+8)(k+1)2
⇔VT=VP⇔VT=VP
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