求连续几个奇数的平方和1的平方+2的平方+3的平方+-------+n的平方=?
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![求连续几个奇数的平方和1的平方+2的平方+3的平方+-------+n的平方=?](/uploads/image/z/6928221-21-1.jpg?t=%E6%B1%82%E8%BF%9E%E7%BB%AD%E5%87%A0%E4%B8%AA%E5%A5%87%E6%95%B0%E7%9A%84%E5%B9%B3%E6%96%B9%E5%92%8C1%E7%9A%84%E5%B9%B3%E6%96%B9%2B2%E7%9A%84%E5%B9%B3%E6%96%B9%2B3%E7%9A%84%E5%B9%B3%E6%96%B9%2B-------%2Bn%E7%9A%84%E5%B9%B3%E6%96%B9%3D%3F)
求连续几个奇数的平方和1的平方+2的平方+3的平方+-------+n的平方=?
求连续几个奇数的平方和
1的平方+2的平方+3的平方+-------+n的平方=?
求连续几个奇数的平方和1的平方+2的平方+3的平方+-------+n的平方=?
=1²+2²+3²+.+n²
=n(n+1)(2n+1)/6
由于1²+2²+3²+.+(2n)²
=2n(2n+1)(4n+1)/6
=n(2n+1)(4n+1)/3
所以n(2n+1)(4n+1)/3=S+2n(n+1)(2n+1)/3
S=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3
=n[(2n+1)(4n+1)-2(n+1)(2n+1)]/3
=n[(2n+1)(4n+1-2n-2)]/3
=n[(2n+1)(2n-1)]/3
=n(4n²-1)/3
懂了吗
1的平方+2的平方+3的平方+-------+n的平方
=1²+2²+3²+......+n²
=n(n+1)(2n+1)/6
求连续几个奇数的平方和
令S=1²+3²+5²+......(2n-1)²
1²+2²+3²+......+(2n)...
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1的平方+2的平方+3的平方+-------+n的平方
=1²+2²+3²+......+n²
=n(n+1)(2n+1)/6
求连续几个奇数的平方和
令S=1²+3²+5²+......(2n-1)²
1²+2²+3²+......+(2n)²
=[1²+3²+5²+......(2n-1)²]+[2²+4²+6²+......+(2n)²]
=[1²+3²+5²+......(2n-1)²]+4[1²+2²+3²+......+n²]
=S+4n(n+1)(2n+1)/6
=S+2n(n+1)(2n+1)/3
由于1²+2²+3²+......+(2n)²
=2n(2n+1)(4n+1)/6
=n(2n+1)(4n+1)/3
所以n(2n+1)(4n+1)/3=S+2n(n+1)(2n+1)/3
S=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3
=n[(2n+1)(4n+1)-2(n+1)(2n+1)]/3
=n[(2n+1)(4n+1-2n-2)]/3
=n[(2n+1)(2n-1)]/3
=n(4n²-1)/3
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