用数学归纳法证明2²;+4²;+6²;+……+(2n)²=1/4n(n+1)(2n+1)
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![用数学归纳法证明2²;+4²;+6²;+……+(2n)²=1/4n(n+1)(2n+1)](/uploads/image/z/1747976-32-6.jpg?t=%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E2%26%23178%3B%3B%2B4%26%23178%3B%3B%2B6%26%23178%3B%3B%2B%E2%80%A6%E2%80%A6%2B%EF%BC%882n%29%26%23178%3B%3D1%2F4n%28n%2B1%29%282n%2B1%29)
用数学归纳法证明2²;+4²;+6²;+……+(2n)²=1/4n(n+1)(2n+1)
用数学归纳法证明2²;+4²;+6²;+……+(2n)²=1/4n(n+1)(2n+1)
用数学归纳法证明2²;+4²;+6²;+……+(2n)²=1/4n(n+1)(2n+1)
正确的结论是:2²+4²+6²+……+﹙2n﹚²=﹙2/3﹚n﹙n+1﹚﹙2n+1)
证明:
1、当n=1时,2^2=2/3*1*2*3,符合题述公式
2、下面证明,当f(n)=2^2+4^2+6^2+...+[2n]^2=2/3*n(n+1)(2n+1)时
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)
综上所述,当f(n)=2^2+4^2+6^2+...+[2n]^2=2/3*n(n+1)(2n+1)时
f(n+1)=2^2+4^2+6^2+...+[2n]^2+[2(n+1)]^2=2/3*(n+1)(n+2)(2n+3)
又因为当n=1时,2^2=2/3*1*2*3,符合题述公式
所以题述公式成立
正确的结论是:2²+4²+6²+……+﹙2n﹚²=﹙2/3﹚n﹙n+1﹚﹙2n+1﹚。 证明:⑴当n=1时,等式左边=2²=4,等式右边=﹙2/3﹚×1×2×3=4,成立。⑵假设当n=k时等式成立,即:2²+4²+……+﹙2k﹚²=﹙2/3﹚k﹙k+1﹚﹙2k+1﹚,则当n=k+1时,等式左边=2...
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正确的结论是:2²+4²+6²+……+﹙2n﹚²=﹙2/3﹚n﹙n+1﹚﹙2n+1﹚。 证明:⑴当n=1时,等式左边=2²=4,等式右边=﹙2/3﹚×1×2×3=4,成立。⑵假设当n=k时等式成立,即:2²+4²+……+﹙2k﹚²=﹙2/3﹚k﹙k+1﹚﹙2k+1﹚,则当n=k+1时,等式左边=2²+4²+……+﹙2k﹚²+﹙2k+2﹚²=﹙2/3﹚k﹙k+1﹚﹙2k+1﹚+﹙2k+2﹚²=﹙k+1﹚[﹙2/3﹚k﹙2k+1﹚+4﹙k+1﹚]=﹙2/3﹚﹙k+1﹚[2k²+k+6k+6]=﹙2/3﹚﹙k+1﹚[﹙k+1﹚+1][2﹙k+1﹚+1],即当n=k+1时,等式成立,∴原式成立。
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证明n(n+1)(2n+1)/4 + (2(n+1))^2 = (n+1)(n+2)(2n+3)/4