1+2的平方+3的平方+……+n的平方怎么算
4个回答

由1²+2²+3²+.+n²=n(n+1)(2n+1)/6

∵(a+1)³-a³=3a²+3a+1(即(a+1)³=a³+3a²+3a+1)

a=1时:2³-1³=3×1²+3×1+1

a=2时:3³-2³=3×2²+3×2+1

a=3时:4³-3³=3×3²+3×3+1

a=4时:5³-4³=3×4²+3×4+1

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a=n时:(n+1)³-n³=3×n²+3×n+1

等式两边相加:

(n+1)³-1=3(1²+2²+3²+.+n²)+3(1+2+3+.+n)+(1+1+1+.+1)

3(1²+2²+3²+.+n²)=(n+1)³-1-3(1+2+3+.+n)-(1+1+1+.+1)

3(1²+2²+3²+.+n²)=(n+1)³-1-3(1+n)×n÷2-n

6(1²+2²+3²+.+n²)=2(n+1)³-3n(1+n)-2(n+1)

=(n+1)[2(n+1)²-3n-2]

=(n+1)[2(n+1)-1][(n+1)-1]

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

∴1²+2²+.+n²=n(n+1)(2n+1)/6.