试证明:1²+2²+3²+……+N²=1/6N(N+1)(2N+1)
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试证明:1²+2²+3²+……+N²=1/6N(N+1)(2N+1)
(n+1)³-n³=3n²+3n+1
故
2³-1³=3×1²+3×1+1
3³-2³=3×2²+3×2+1
4³-3³=3×3²+3×3+1
……
(n+1)³-n³=3n²+3n+1
由等式的叠加性可知,左边相加=右边相加,即
(n+1)³-1=3(1²+2²+3²+……+n²)+3(1+2+3+……+n)+n
n³+3n²+3n+1-1=3(1²+2²+3²+……+n²)+3×1/2n(n+1)+n
3(1²+2²+3²+……+n²)=n³+3n²+3n-1.5n²-2.5n=n³+1.5n²+0.5n=1/2n×(2n²+3n+1)
=1/2n(n+1)(2n+1)
故1²+2²+3²+……+n²=1/6n(n+1)(2n+1)
故
2³-1³=3×1²+3×1+1
3³-2³=3×2²+3×2+1
4³-3³=3×3²+3×3+1
……
(n+1)³-n³=3n²+3n+1
由等式的叠加性可知,左边相加=右边相加,即
(n+1)³-1=3(1²+2²+3²+……+n²)+3(1+2+3+……+n)+n
n³+3n²+3n+1-1=3(1²+2²+3²+……+n²)+3×1/2n(n+1)+n
3(1²+2²+3²+……+n²)=n³+3n²+3n-1.5n²-2.5n=n³+1.5n²+0.5n=1/2n×(2n²+3n+1)
=1/2n(n+1)(2n+1)
故1²+2²+3²+……+n²=1/6n(n+1)(2n+1)
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