1的平方一直加到N的平方的计算公式？ 1的平方+3的平方。。。一直加到(2n-1)平方
1\u7684\u5e73\u65b9\u52a02\u7684\u5e73\u65b9\u4e00\u76f4\u52a0\u5230n\u7684\u5e73\u65b9\u7b49\u4e8e\u591a\u5c11\u7ed3\u679c\u4e3a\uff1an\uff082n-1\uff09\uff082n+1\uff09/3
\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u6269\u5c55\u8d44\u6599\u6570\u5217\u6027\u8d28\uff1a
\uff081\uff09\u901a\u9879\u516c\u5f0f\uff1aan=a1+(n-1)d
\uff082\uff09\u901a\u9879\u516c\u5f0f\u7684\u63a8\u5e7f\uff1a\u4efb\u610f\u4e24\u9879
,
\u7684\u5173\u7cfb\u4e3a
=
\uff083\uff09\u4ece\u7b49\u5dee\u6570\u5217\u7684\u5b9a\u4e49\u3001\u901a\u9879\u516c\u5f0f\u3001\u524dn\u9879\u548c\u516c\u5f0f\u53ef\u4ee5\u63a8\u51fa:
\uff0ck\u2208{1,2,\u2026,n}
\uff084\uff09\u82e5m\uff0cn\uff0cp\uff0cq\u2208N*\uff0c\u4e14m+n=p+q\uff0c\u5219\u6709am+an=ap+aq
\uff085\uff09\u82e5m\uff0cn\uff0cp\u2208N*\uff0c\u4e14m+n=2p\uff0c\u5219\u6709am+an=2ap
\uff086\uff09\u7b49\u5dee\u4e2d\u9879\u516c\u5f0f\uff1a\u82e5
\u6210\u7b49\u5dee\u6570\u5217\uff0c\u5219\u6709
\uff087\uff09\u524dn\u9879\u548c\u516c\u5f0f\u4e3a\uff1aSn=na1+[n(n-1)/2] d\u6216Sn=(a1+an)n/2
\u8ba1\u7b97\u65b9\u6cd5\uff1a
\u901a\u9879\u516c\u5f0f an=a1*q^(n-1)(\u5373q\u7684n-1\u6b21\u65b9) a1\u4e3a\u9996\u9879,an\u4e3a\u7b2cn\u9879\uff0can=a1*q^(n-1),am=a1*q^(m-1)\uff0c\u5219an/am=q^(n-m)\u3002
(1)an=am*q^(n-m)
(2)a,G,b \u82e5\u6784\u6210\u7b49\u6bd4\u4e2d\u9879,\u5219G^2=ab (a,b,G\u4e0d\u7b49\u4e8e0)
(3)\u82e5m+n=p+q \u5219 am\u00d7an=ap\u00d7aq
2.\u7b49\u6bd4\u6570\u5217\u524dn\u9879\u548c
\u8bbe a1,a2,a3...an\u6784\u6210\u7b49\u6bd4\u6570\u5217\uff0c\u524dn\u9879\u548cSn=a1+a2+a3...an
Sn=a1+a1*q+a1*q^2+....a1*q^(n-2)+a1*q^(n-1)(\u8fd9\u4e2a\u516c\u5f0f\u867d\u7136\u662f\u6700\u57fa\u672c\u516c\u5f0f,\u4f46\u4e00\u90e8\u5206\u9898\u76ee\u4e2d\u6c42\u524dn\u9879\u548c\u662f\u5f88\u96be\u7528\u4e0b\u9762\u90a3\u4e2a\u516c\u5f0f\u63a8\u5bfc\u7684,\u8fd9\u65f6\u53ef\u80fd\u8981\u76f4\u63a5\u4ece\u57fa\u672c\u516c\u5f0f\u63a8\u5bfc\u8fc7\u53bb,\u6240\u4ee5\u5e0c\u671b\u8fd9\u4e2a\u516c\u5f0f\u4e5f\u8981\u7406\u89e3)
Sn=a1(1-q^n)/(1-q)=(a1-an*q)/(1-q)
具体算法
利用立方差公式
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
各等式全相加就得到咯。
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