请计算1的平方加二的平方加三的平方一直加到1999的平方，结果是多少，要过程
\u4e00\u7684\u5e73\u65b9\u52a0\u4e8c\u7684\u5e73\u65b9\u52a0\u4e09\u7684\u5e73\u65b9\u4e00\u76f4\u52a0\u5230\u4e00\u767e\u7684\u5e73\u65b9\u7b49\u4e8e\u591a\u5c11\u89e3\uff1a\u516c\u5f0f\uff1a
1²+2²+3²+....+n²
=1/6 n(n+1)(2n+1)
\u6240\u4ee5
\u53d6n=100\uff0c\u5f97
\u539f\u5f0f=1/6 \u00d7100\u00d7\uff08100+1\uff09\u00d7\uff082\u00d7100+1\uff09
=338350
\u4e00\u7684\u5e73\u65b9\u52a0\u4e8c\u7684\u5e73\u65b9\u52a0\u4e09\u7684\u5e73\u65b9\u00b7\u00b7\u00b7\u00b7\u00b7\u4e00\u76f4\u52a0\u5230n\u7684\u5e73\u65b9
=n(n+1)(2n+1)/6
n=9
9\u00d7(9+1)\u00d7(2\u00d79+1)\u00f76=285
1^2+2^2+3^2+.....+n^2=1/6 n(n+1)(2n+1)
证明过程略
带入a=1999
可得原式=1/6*(1999+1)(3998+1)
=1/6*799800
=1333000
1^2+2^2+3^2+......+1999^2
=(1x2-1)+(2x3-2)+(3x4-3)+......+(1999x2000-1999)
=(1x2+2x3+3x4+......+1999x2000)-(1+2+3+......+1999)
=[(1x2x3-0x1x2)+(2x3x4-1x2x3)+(3x4x5-2x3x4)+......+(1999x2000x2001-1998x1999x2000)]÷3-(1+2+3+......+1999)
=(1999x2000x2001-0x1x2)÷3-(1+1999)x1999÷2
=1999x2000x2001÷3-1999000
=2666666000-1999000
=2664667000
http://wapbaike.baidu.com/view/892600.htm?ref=wise&bd_source_light=1
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绛旓細1^2+2^2+3^2+...+n^2=1/6 n(n+1)(2n+1)璇佹槑杩囩▼鐣 甯﹀叆a=1999 鍙緱鍘熷紡=1/6*(1999+1)(3998+1)=1/6*799800 =1333000
绛旓細1^2+2^2+3^2 =1+4+9 =14
绛旓細鍥炵瓟锛氫唬鍏(n+1)(2n+1)/6
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绛旓細涓鐨勫钩鏂瑰姞浜岀殑骞虫柟鍔犱笁鐨骞虫柟路路路涓鐩村姞鍒皀鐨勫钩鏂 =n(n+1)(2n+1)/6
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