求1的平方加到n的平方的推导公式，求精确详细，格式完整，尽量简单易懂不繁复。 1的平方加2的平方一直加到n的平方公式如何推导
1\u7684\u5e73\u65b9\u52a0\u5230n\u7684\u5e73\u65b9\u600e\u4e48\u7b97\uff0c\u7528\u6570\u5217\u7684\u65b9\u6cd5\u8fd9\u4e2a\u6709\u8457\u540d\u7684 \u63a8\u5bfc\u516c\u5f0f
Sn=n(n+1)(2n+1)/6
\u63a8\u5bfc\u7684\u8fc7\u7a0b\u8981\u7528\u6570\u5b66\u5f52\u7eb3\u6cd5
\u8fd9\u4e2a\u516c\u5f0f\u8bb0\u4f4f\u5373\u53ef \u8981\u8bc1\u660e\u548c\u53d1\u73b0\u7684\u8bdd \u662f\u4e2a\u5f88\u7e41\u7410\u7684\u8fc7\u7a0b
\u5982\u679c\u6709\u5174\u8da3\u7684\u8bdd \u4f60\u53ef\u4ee5\u770b\u8fd9\u4e2a\u63a8\u5bfc\u8fc7\u7a0b
http://wenku.baidu.com/link?url=9XqMICKdNpj3Tg7DwBW34rdeuS202AwZBvvJQikA6qJIbEAEozN6WTD_srdMqEIXOX60ByKWAr_vbWRErV1EeIGdkxKkdBivLNz3KG1yGu3
\u75311²+2²+3²+\u3002\u3002\u3002+n²=n\uff08n+1)\uff082n+1\uff09/6
\u2235\uff08a+1\uff09³-a³=3a²+3a+1\uff08\u5373\uff08a+1)³=a³+3a²+3a+1\uff09
a=1\u65f6\uff1a2³-1³=3\u00d71²+3\u00d71+1
a=2\u65f6\uff1a3³-2³=3\u00d72²+3\u00d72+1
a=3\u65f6\uff1a4³-3³=3\u00d73²+3\u00d73+1
a=4\u65f6\uff1a5³-4³=3\u00d74²+3\u00d74+1
\u3002\u3002\u3002\u3002\u3002\u3002
a=n\u65f6\uff1a\uff08n+1\uff09³-n³=3\u00d7n²+3\u00d7n+1
\u7b49\u5f0f\u4e24\u8fb9\u76f8\u52a0\uff1a
\uff08n+1)³-1=3\uff081²+2²+3²+\u3002\u3002\u3002+n²\uff09+3\uff081+2+3+\u3002\u3002\u3002+n\uff09+\uff081+1+1+\u3002\u3002\u3002+1\uff09
3\uff081²+2²+3²+\u3002\u3002\u3002+n²\uff09=\uff08n+1\uff09³-1-3\uff081+2+3+\u3002\u3002\u3002+n\uff09-\uff081+1+1+\u3002\u3002\u3002+1\uff09
3\uff081²+2²+3²+\u3002\u3002\u3002+n²\uff09=\uff08n+1\uff09³-1-3\uff081+n\uff09\u00d7n\u00f72-n
6\uff081²+2²+3²+\u3002\u3002\u3002+n²\uff09=2\uff08n+1)³-3n\uff081+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)
\u22341²+2²+\u3002\u3002\u3002+n²=n\uff08n+1)\uff082n+1\uff09/6.
∵（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
。。。。。。
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-n6（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.
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