Ta có công thức:
\(1^2+2^2+3^2+...+n^2=\dfrac{n.\left(n+1\right).\left(2n+1\right)}{6}\left(1\right)\)
Đặt:
\(D.(2n + 1) = 1^2 + 3^2 + 5^2 + ... + (2n+1)^2 \)
\(D(2n) = 2^2 + 4^2 + ...+ (2n)^2 \)
\(= 2^2.(1^2 + 2^2+ 3^2 +...+ n^2) \)
\(=\dfrac{4n.(n+1)(2n+1) }{6} \)
\(\Rightarrow D(2n+1) + D(2n)\)
\(= 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + ... + (2n)^2 + (2n+1)^2 \)
Áp dụng (1) ta có:
\(D(2n+1) + D(2n)\)
\(=(2n+1).[(2n+1) + 1).[2(2n+1) + 1)]/6\)
\(= (2n+1).(2n+2).(4n+3)/6 \)
\(\Rightarrow D(2n+1)\)
\(= (2n+1).(2n+2).(4n+3)/6 - D(2n) \)
\(= 2(2n+1).(n+1).(4n+3)/6 - 4n.(n+1).(2n+1)/6 \)
\(= (2n+1).(n+1)/3.[(4n +3) - 2n] = (2n+1).(n+1).(2n+3)/3 \)
Vậy \(D(2n+1)=(2n+1).(n+1).\left(2.n+3\right)\text{/}3\)
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