请问仪的平方+二的平方+三的平方+.+n的平方,怎么计算,主要要计算过程,

1个回答

  • 因为:(n+1)³=n³+3n²+3n+1

    则有: 1³=1³

    2³=1³+3×1²+3×1+1

    3³=2³+3×2²+3×2+1

    ……

    n³=(n-1)³+3(n-1)²+3(n-1)+1

    (n+1)³=n³+3n²+3n+1

    上述各式相加得:(n+1)³=1³+3×(1²+2²+3²+……+n²)+3×(1+2+3+……+n)+n

    3×(1²+2²+3²+……+n²)

    =(n+1)³-n-1-3×(1+2+3+……+n)

    =n³+3n²+2n-3/2×n(n+1)

    =n³+3n²/2+n/2

    =1/2×(2n³+3n²+n)

    =n(2n+1)(n+1)/2

    所以:1²+2²+3²+……+n²=n(2n+1)(n+1)/6