1平方加到n平方简算过程及证明方法 1的平方加到n的平方怎么算，用数列的方法

1\u7684\u5e73\u65b9\u52a02\u7684\u5e73\u65b9\u4e00\u76f4\u52a0\u5230n\u7684\u5e73\u65b9\u516c\u5f0f\u5982\u4f55\u63a8\u5bfc

\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.

\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

因为（k+1）^3=k^3+3k^2+3k+1......1
k^3=(k-1)^3+3(k-1)^2+3(k-1)+1.....2
........
........
2^3=1^3+3*1^2+3*1+1......k
k式相加：(k+1)^3-1=3(k^2+....+1)+3(k+k-1+....+1)+k
所以3（k^2+...+1)
=(k+1)[(k+1)^2-1-k-(3k(k+1)/2)]
=k(k+1)(2k+1)
故1^2+2^2+3^2+...+N^2=N(N+1)(2N+1)/6

(n+1)^3-n^3=3n^2+3n+12^3-1^3=3*1^2+3*1+1
3^3-2^3=3*2^2+3*2+1
4^3-3^3=3*3^2+3*3+1
......(n+1)^3-n^3=3n^2+3n+1
以上各式相加,可得:(n+1)^3-1^3=3(
1^2+2^2+3^2+...+n^2)+3(1+2+3+4+5+6+.....+n)+n
即n^3+3n^2+3n=3(1^2+2^2+3^2+...+n^2)+3n(n+1)/2+n
整理即可得1^2+2^2+3^2+...+N^2=N(N+1)(2N+1)/6

求1^2+2^2+....+n^2=?
因为有
(n+1)^3-n^3=3n^2+3n+1
n^3-(n-1)^3=3(n-1)^2+3(n-1)+1
……
2^3-1^3=3*1^2+3*1+1
以上所有式等号左右分别相加,有
(n+1)^3-1=3*[n^2+(n-1)^2+(n-2)^2+……+2^2+1^2]+3*[n+(n-1)+……+3+2+1]+n*1
并且因为有
(n+1)^3-1=(n+1-1)[(n+1)^2+(n+1)+1]
=n(n^2+3n+3)
n+(n-1)+……+3+2+1=n(n+1)/2
所以,
n^2+(n-1)^2+(n-2)^2+……+2^2+1^2={[(n+1)^3-1]-3*[n+(n-1)+……+3+2+1]-n*1}/3
=[n(n^2+3n+3)-3n(n+1)/2-n]/3
=[n(n^2+3n+3-3n/2-3/2-1)]/3
=n(2n^2+3n+1)/6
=n(n+1)(2n+1)/6

2鐨勫钩鏂+4鐨勫钩鏂+6鐨勫钩鏂+8鐨勫钩鏂逛竴鐩鍔犲埌2004鐨勫钩鏂规庢牱绠绠?
绛旓細2^2+4^2+...+2004^2 =(2*1)^2+(2*2)^2+...+(2*1002)^2 =2^2*(1^2+2^2+...+1002^2)=4*1002*(1002+1)*(2*1002+1)/6 =1343358020 鍏紡锛 1^2+2^2+...n^2=n(n+1)(2n+1)/6

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绛旓細1-2+3-4+5-6+鈥︹︹-100+101 =锛1+2锛夛紙1-2锛+锛3+4锛夛紙3-4锛+鈥︹+锛99+100锛夛紙99-100锛+101 =-锛1+2锛-锛3+4锛-鈥︹︹旓紙99+100锛+101 =锛-1锛壝楋紙1+2+3+鈥︹+100锛+101 =101-5050 =10201-5050 =5151

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