12345…n的平方和 1+2的平方+3的平方+ +n的平方,和怎么计算

\u75af\u72c2\u6570\u5b66\uff1a1\u30012\u30013\u30014\u30015\u3001\u2026\u2026\u202619856\u7684\u5e73\u65b9\u548c\u662f\u51e0\uff1f 1\u30012\u30013\u30014\u30015\u3001\u2026\u2026\u2026n\u7684\u5e73\u65b9\u548c\u662f\u51e0\uff1f

\u5e73\u65b9\u548c\u516c\u5f0fn(n+1)(2n+1)/6 \u3000\u3000
\u53731^2+2^2+3^2+\u2026+n^2=n(n+1)(2n+1)/6 (\u6ce8\uff1an^2=n\u7684\u5e73\u65b9)
\u6839\u636e\u516c\u5f0f\uff0c1\u30012\u30013\u30014\u30015\u3001\u2026\u2026\u202619856\u7684\u5e73\u65b9\u548c\u662f2609677525016

1²+2²+3²+...+n²=n(n+1)(2n+1)/6.
\u63a8\u7406\u5982\u4e0b:
2³-1³=3\u00d71²+3\u00d71+1
3³-2³=3\u00d72²+3\u00d72+1
4³-3³=3\u00d73²+3\u00d72+1
... ...
(n+1)³-n³=3n²+3n+1
\u4ee5\u4e0an\u4e2a\u5f0f\u5b50\u76f8\u52a0\uff0c\u5f97
(n+1)³-1=3(1²+2²+3²+...+n²)+3(1+2+3+...+n)+(1+1+1+...+1)
\u5373(n+1)³-1=3(1²+2²+3²+...+n²)+3[n(n+1)/2]+n
\u22343S=(n+1)³-1-3n(n+1)/2-(n+1)
\u5373S=n(n+1)(2n+1)/6\u3002
\u6269\u5c55\u8d44\u6599
\u5e73\u65b9\u548c\u76f8\u5173\u516c\u5f0f\uff1a
\uff081\uff091+2+3+.+n=n(n+1)/2
\uff082\uff091^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
\uff083\uff091\u00d72\uff0b2\u00d73\uff0b3\u00d74\uff0b4\u00d75\uff0b\u2026\uff0bn(n\uff0b1)
\uff1d(1^2+1)+(2^2+2)+(3^2+2)+...+(n^2+n)
=(1^2+2^2+...+n^2)+(1+2+3+.+n)
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)

1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6.

n×(n+1)x(2n+1)
÷6

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