正確的結(jié)論是：2²＋4²＋6²＋……＋﹙2n﹚²=﹙2／3﹚n﹙n＋1﹚﹙2n＋1）
證明：
1、當(dāng)n=1時,2^2=2/3*1*2*3,符合題述公式
2、下面證明,當(dāng)f(n)=2^2+4^2+6^2+...+[2n]^2=2/3*n(n+1)(2n+1)時
f(n+1)=2^2+4^2+6^2+...+[2n]^2+[2(n+1)]^2=2/3*(n+1)(n+2)(2n+3)
f(n+1)=f(n)+[2(n+1)]^2
=2/3*n(n+1)(2n+1)+[2(n+1)]^2
=[2n(n+1)(2n+1)+12(n+1)(n+1)]/3
=[4n^2+2n+12n+12](n+1)/3
=[4n^2+14n+12](n+1)/3
=2[2n^2+7n+6](n+1)/3
=2(2n+3)(n+2)(n+1)/3
=2/3*(n+1)(n+2)(2n+3)
綜上所述,當(dāng)f(n)=2^2+4^2+6^2+...+[2n]^2=2/3*n(n+1)(2n+1)時
f(n+1)=2^2+4^2+6^2+...+[2n]^2+[2(n+1)]^2=2/3*(n+1)(n+2)(2n+3)
又因為當(dāng)n=1時,2^2=2/3*1*2*3,符合題述公式
所以題述公式成立