1^2+2^2+3^2+…+(2n)^2=1/3n(2n+1)(4n+1) 用数学归纳法证明.

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1^2+2^2+3^2+…+(2n)^2=1/3n(2n+1)(4n+1)用数学归纳法证明.1^2+2^2+3^2+…+(2n)^2=1/3n(2n+1)(4n+1)用数学归纳法证明.1^2+2^2+

1^2+2^2+3^2+…+(2n)^2=1/3n(2n+1)(4n+1) 用数学归纳法证明.
1^2+2^2+3^2+…+(2n)^2=1/3n(2n+1)(4n+1) 用数学归纳法证明.

1^2+2^2+3^2+…+(2n)^2=1/3n(2n+1)(4n+1) 用数学归纳法证明.
1)n=1时,左=1^2+2^2=5,右=1/3*1*3*5=5,左=右,命题成立.
2)设当n=k时,命题成立,即
1^2+2^2+3^2+...+(2k)^2=1/3*k(2k+1)(4k+1)
则当n=k+1时,
1^2+2^2+3^2+...+(2k)^2+(2k+1)^2+(2k+2)^2
=1/3*k(2k+1)(4k+1)+(2k+1)^2+(2k+2)^2
=1/3(2k+1)*[k(4k+1)+3(2k+1)]+4(k+1)^2
=1/3(2k+1)*(4k^2+7k+3)+4(k+1)^2
=1/3(2k+1)(k+1)(4k+3)+4(k+1)^2
=1/3(k+1)*[(2k+1)(4k+3)+12(k+1)]
=1/3(k+1)(8k^2+22k+15)
=1/3(k+1)(2k+3)(4k+5)
就是说,当n=k+1时,命题也成立.
根据1)、2)可知,命题对所有正整数n都成立.

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