1的平方加到n的平方怎么算，用数列的方法 1的平方加到100的平方怎么算
\u6c421\u7684\u5e73\u65b9\u52a0\u4e0a2\u7684\u5e73\u65b9 \u4e00\u76f4\u52a0\u5230n\u7684\u5e73\u65b9\u516c\u5f0f\u7684\u63a8\u5bfc\u8fc7\u7a0b \u5199\u5728\u7eb8\u4e0a \u8fd0\u7528\u6570\u5217\u76f8\u5173\u77e5\u8bc6\u8bc1\u6cd5\u4e00
n^2=n(n+1)-n
1^2+2^2+3^2+.+n^2
=1*2-1+2*3-2+.+n(n+1)-n
=1*2+2*3+...+n(n+1)-(1+2+...+n)
\u7531\u4e8en(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
\u6240\u4ee51*2+2*3+...+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+.+n(n+1)(n+2)-(n-1)n(n+1)]/3
[\u524d\u540e\u6d88\u9879]
=[n(n+1)(n+2)]/3
\u6240\u4ee51^2+2^2+3^2+.+n^2
=[n(n+1)(n+2)]/3-[n(n+1)]/2
=n(n+1)[(n+2)/3-1/2]
=n(n+1)[(2n+1)/6]
=n(n+1)(2n+1)/6
\u8bc1\u6cd5\u4e8c
\u5229\u7528\u7acb\u65b9\u5dee\u516c\u5f0f
n^3-(n-1)^3
=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n
=2*n^2+(n-1)^2-n
2^3-1^3=2*2^2+1^2-2
3^3-2^3=2*3^2+2^2-3
4^3-3^3=2*4^2+3^2-4
.
n^3-(n-1)^3=2*n^2+(n-1)^2-n
\u5404\u7b49\u5f0f\u5168\u90e8\u76f8\u52a0
n^3-1^3=2*(2^2+3^2+...+n^2)+[1^2+2^2+...+(n-1)^2]-(2+3+4+...+n)
n^3-1=2*(1^2+2^2+3^2+...+n^2)-2+[1^2+2^2+...+(n-1)^2+n^2]-n^2-(2+3+4+...+n)
n^3-1=3*(1^2+2^2+3^2+...+n^2)-2-n^2-(1+2+3+...+n)+1
n^3-1=3(1^2+2^2+...+n^2)-1-n^2-n(n+1)/2
3(1^2+2^2+...+n^2)
=n^3+n^2+n(n+1)/2
=(n/2)(2n^2+2n+n+1)
=(n/2)(n+1)(2n+1)
1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
\u5957\u7528\u516c\u5f0f\uff1a1^2+2^2+3^2+4^2+5^2\u2026\u2026\u2026\u2026\u2026\u2026+n^2=n(n+1)(2n+1)/6\u3002
1^2+2^2+3^2+4^2+5^2\u2026\u2026\u2026\u2026\u2026\u2026+100^2=100\u00d7101\u00d7201\u00f76=338 350\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u5e38\u7528\u5e73\u65b9\u6570\uff1a
1² = 1\uff0c 2² = 4 \uff0c3² = 9\uff0c 4² = 16\uff0c 5² = 25\uff0c 6² = 36 \uff0c7² = 49 \uff0c8² = 64 \uff0c9² = 81 \uff0c10² = 100
11² = 121\uff0c 12² = 144 \uff0c13² = 169 \uff0c14² = 196 \uff0c15² = 225\uff0c 16² = 256\uff0c 17² = 289 \uff0c18² = 324\uff0c 19² = 361 \uff0c20² = 400
21² = 441 \uff0c22² = 484\uff0c 23² = 529 \uff0c24² = 576\uff0c 25² = 625 \uff0c26² = 676\uff0c 27² = 729 \uff0c28² = 784 \uff0c29² = 841\uff0c 30² = 900
\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)
Sn=n(n+1)(2n+1)/6
推导的过程要用数学归纳法
这个公式记住即可 要证明和发现的话 是个很繁琐的过程
如果有兴趣的话 你可以看这个推导过程
http://wenku.baidu.com/link?url=9XqMICKdNpj3Tg7DwBW34rdeuS202AwZBvvJQikA6qJIbEAEozN6WTD_srdMqEIXOX60ByKWAr_vbWRErV1EeIGdkxKkdBivLNz3KG1yGu3
你可以看一下这个方法
1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6，平方和公式，可以用数学归纳法证
如果需要自己推导，需要借助于差分，是一个较繁琐的过程。
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