1的平方加2的平方加3的平方一直加到100的平方是多少? 1平方加2平方加3平方一直加到n平方等于多少
1\u7684\u5e73\u65b9+2\u7684\u5e73\u65b9+3\u7684\u5e73\u65b9+.+99\u7684\u5e73\u65b9\u9664\u4ee54\u7684\u4f59\u6570\u662f\u591a\u5c11\u8003\u8651\u6bcf\u4e2a\u5e73\u65b9\u7684\u5bf94\u7684\u4f59\u6570
1^1 = 1 =>1
2^2 = 4 =>0
3^2 = 9 =>1
4^2 =16 =>0
5^2 = 25=>1
6^2 = 36=>0
7^2 = 49=>1
8^2=64=>0
9^2=>81=>1
10^2 =100=>0
11^2=121=>1
12^2 = 144=>0
13^2=169=>1
.
\u53d1\u73b0\u4e2a\u4f4d\u6570\u662f\u5faa\u73af\u51fa\u73b0\u7684
\u90a3\u4e48\u52a0\u523099^2\u7684\u65f6\u5019
\u6700\u540e\u5bf94\u7684\u4f59\u6570\u8981\u7b49\u4e8e (99+1)/2 = 100/2 = 50
\u800c50\u5bf94\u7684\u4f59\u6570\u662f2
\u90a3\u4e481\u7684\u5e73\u65b9+2\u7684\u5e73\u65b9+\u2026\u2026+99\u7684\u5e73\u65b9\u00f74 \u4f59\u6570\u662f 2
1²+2²+3²+\u2026\u2026+n²=n(n+1)(2n+1)/6\u3002\u53ef\u4ee5\u7528(n+1)³-n³=3n²+3n+1\u7d2f\u52a0\u5f97\u5230\u3002
\u8bc1\u660e\u8fc7\u7a0b\uff1a
\u6839\u636e\u7acb\u65b9\u5dee\u516c\u5f0f(a+1)³-a³=3a²+3a+1\uff0c\u5219\u6709\uff1a
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.\u00b7\u00b7
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+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+n²\uff09+3\uff081+2+3+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+n\uff09+\uff081+1+1+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+1\uff09
3\uff081²+2²+3²+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+n²\uff09=\uff08n+1\uff09³-1-3\uff081+2+3+.+n\uff09-\uff081+1+1+.+1\uff09
3\uff081²+2²+3²+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+n²\uff09=\uff08n+1\uff09³-1-3\uff081+n\uff09\u00d7n\u00f72-n
6\uff081²+2²+3²+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+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)
\u6240\u4ee51²+2²+\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7+n²=n\uff08n+1)\uff082n+1\uff09/6\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u7acb\u65b9\u5dee\u516c\u5f0f\u4e0e\u7acb\u65b9\u548c\u516c\u5f0f\u7edf\u79f0\u4e3a\u7acb\u65b9\u516c\u5f0f\uff0c\u4e24\u8005\u57fa\u672c\u63cf\u8ff0\u5982\u4e0b:
1\u3001\u7acb\u65b9\u548c\u516c\u5f0f\uff0c\u5373\u4e24\u6570\u7acb\u65b9\u548c\u7b49\u4e8e\u8fd9\u4e24\u6570\u7684\u548c\u4e0e\u8fd9\u4e24\u6570\u5e73\u65b9\u548c\u4e0e\u8fd9\u4e24\u6570\u79ef\u7684\u5dee\u7684\u79ef\u3002\u4e5f\u53ef\u4ee5\u8bf4\u4e24\u6570\u7acb\u65b9\u548c\u7b49\u4e8e\u8fd9\u4e24\u6570\u79ef\u4e0e\u8fd9\u4e24\u6570\u5dee\u7684\u4e0d\u5b8c\u5168\u5e73\u65b9\u7684\u79ef\u3002
2\u3001\u7acb\u65b9\u5dee\u516c\u5f0f\uff0c\u5373\u4e24\u6570\u7acb\u65b9\u5dee\u7b49\u4e8e\u8fd9\u4e24\u6570\u5dee\u4e0e\u8fd9\u4e24\u6570\u5e73\u65b9\u548c\u4e0e\u8fd9\u4e24\u6570\u79ef\u7684\u548c\u7684\u79ef\u3002\u4e5f\u53ef\u4ee5\u8bf4\uff0c\u4e24\u6570\u7acb\u65b9\u5dee\u7b49\u4e8e\u4e24\u6570\u5dee\u4e0e\u8fd9\u4e24\u6570\u548c\u7684\u4e0d\u5b8c\u5168\u5e73\u65b9\u7684\u79ef \u3002
\u53c2\u8003\u8d44\u6599\uff1a\u767e\u5ea6\u767e\u79d1_\u7acb\u65b9\u5dee\u516c\u5f0f
=100[100+1][2*100+1]/6
=338350
S（n)=n(n+1)(2n+1)/6
1*1+2*2+3*3+4*4+...n*n=n(n+1)(2n+1)/6,
故,本题=338350
338350
绛旓細1^2+2^2+...+n^2=n(n+1)(2n+1)/6
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绛旓細12+22+32+鈥+n2=n(n+1)(2n+1)/6,鍦ㄩ珮涓暟瀛︿腑鏄敤鏁板褰掔撼娉曡瘉鏄庣殑涓涓懡棰,娌℃湁缁欏嚭鍏剁洿鎺ョ殑鎺ㄥ杩囩▼.鍏跺疄,璇ユ眰鍜屽叕寮忕殑鐩存帴鎺ㄥ骞朵笉澶嶆潅,涔熸病鏈夎秴鍑哄垵涓暟瀛﹀唴瀹.璁撅細S=12+22+32+鈥+n2 鍙﹁锛歋1=12+22+32+鈥+n2+(n+1)2+(n+2)2+(n+3)2+鈥+(n+n)2,姝ゆ璁鹃鏄В棰樼殑...
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绛旓細1²+2²+3²+4²+...+n²=n(n+1)(2n+1)/6 鍙敤鏁板褰掔撼娉曡瘉鏄庛
绛旓細鍒╃敤鍏紡锛1^2+2^2+3^3+...+n^2=[n(n+1)(2n+1)]/6 1+2+3+...+n=[n(n+1)]/2 灏卞彲浠ユ眰鍑烘潵浜嗐備竴鐨勫钩鏂瑰姞浜岀殑骞虫柟鍔犱笁鐨勫钩鏂涓鐩村姞鍒颁簩闆1涓冪殑骞虫柟闄や互涓鍔浜屽姞涓変竴鐩村姞鍒2017 =[(2017x2018x4035)/6]/[2017x2018)/2]=4035x2/6 =1345銆
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