归纳法证明1^2+2^2+3^2+……+r^2=r(r+1)(2r+1)/61^2+2^2+3^2+……+r^2+r^2=r(r+1)(2r+1)/6+(r+1)^2这一步怎么得出等于(r+1)(r+2)[2(r+1)+1]/6前面是+(r+1)^2
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![归纳法证明1^2+2^2+3^2+……+r^2=r(r+1)(2r+1)/61^2+2^2+3^2+……+r^2+r^2=r(r+1)(2r+1)/6+(r+1)^2这一步怎么得出等于(r+1)(r+2)[2(r+1)+1]/6前面是+(r+1)^2](/uploads/image/z/8710072-16-2.jpg?t=%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E1%5E2%2B2%5E2%2B3%5E2%2B%E2%80%A6%E2%80%A6%2Br%5E2%3Dr%28r%2B1%29%282r%2B1%29%2F61%5E2%2B2%5E2%2B3%5E2%2B%E2%80%A6%E2%80%A6%2Br%5E2%2Br%5E2%3Dr%28r%2B1%29%282r%2B1%29%2F6%2B%28r%2B1%29%5E2%E8%BF%99%E4%B8%80%E6%AD%A5%E6%80%8E%E4%B9%88%E5%BE%97%E5%87%BA%E7%AD%89%E4%BA%8E%28r%2B1%29%28r%2B2%29%5B2%28r%2B1%29%2B1%5D%2F6%E5%89%8D%E9%9D%A2%E6%98%AF%2B%28r%2B1%29%5E2)
归纳法证明1^2+2^2+3^2+……+r^2=r(r+1)(2r+1)/61^2+2^2+3^2+……+r^2+r^2=r(r+1)(2r+1)/6+(r+1)^2这一步怎么得出等于(r+1)(r+2)[2(r+1)+1]/6前面是+(r+1)^2
归纳法证明1^2+2^2+3^2+……+r^2=r(r+1)(2r+1)/6
1^2+2^2+3^2+……+r^2+r^2=r(r+1)(2r+1)/6+(r+1)^2
这一步怎么得出等于(r+1)(r+2)[2(r+1)+1]/6
前面是+(r+1)^2
归纳法证明1^2+2^2+3^2+……+r^2=r(r+1)(2r+1)/61^2+2^2+3^2+……+r^2+r^2=r(r+1)(2r+1)/6+(r+1)^2这一步怎么得出等于(r+1)(r+2)[2(r+1)+1]/6前面是+(r+1)^2
=r(r+1)(2r+1)/6+(r+1)^2
=(r+1)[r(2r+1)/6+(r+1) ]
=(r+1)[2r^2+r+6r+6]/6
=(r+1)[2r^2+7r+6]/6
1 2
2 3十字相乘
=(r+1)(r+2)[2(r+1)+1]/6
用归纳法。
1)当n=1时,1^2=1*2*3/6=1,等式成立。
2)假设n=k时,1^2+2^2+3^2......+k^2=k(k+1)(2k+1)/6成立。
那么:
1^2+2^2+3^2......+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)/6*[k(2k+1)+6(k+1)]
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用归纳法。
1)当n=1时,1^2=1*2*3/6=1,等式成立。
2)假设n=k时,1^2+2^2+3^2......+k^2=k(k+1)(2k+1)/6成立。
那么:
1^2+2^2+3^2......+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)/6*[k(2k+1)+6(k+1)]
=(k+1)/6*(k+2)(2k+3)
=(k+1)(k+2)[2(k+1)+1]/6
等式也成立。
3)因为n=1等式成立,所以
1^2+2^2+3^2......+n^2=n(n+1)(2n+1)/6
恒成立
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