1的平方+2的平方+3的平方一直加到（n-1）的平方，如何化简？
\u5316\u7b80\uff1a\uff081\u52a0a-2\u5206\u4e4b3)\u9664\u4ee5a\u5e73\u65b9-4\u5206\u4e4ba+1[1+3/(a-2)]\u9664\u4ee5(a+1)/(a^2-4)
=(a+1)/(a+2)\u9664\u4ee5\uff08a+1\uff09/(a+2)(a-2)
=(a+1)/(a+2)\u4e58\u4ee5(a+2)(a-2)/(a+1)
=a-2
(a²+3)/(a+1)(a-1)-(a+1)²/(a+1)(a-1) -1
=(a²+3-a²-2a-1)/(a+1)(a-1)-1
=(2-2a)/(a+1)(a-1)-1
=-2/(a+1)-1
=(-2-a-1)/(a+1)
=-(3+a)/(a+1)
把下面的n替换为n-1即可
网上很多，请参考：
http://zhidao.baidu.com/link?url=ZHh6uPfcaAQh4jmEyrwk0kX4xdL4Ll_z-38wlLh_eKb0ku6NixHIZUL5t8Qh2eNyw6bnpfGAoznKJfLiF5cs6q
1^2+2^2+3^2+4^2+5^2………………+n^2=n(n+1)(2n+1)/6
数学归纳法可以证
也可以如下做 比较有技巧性
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)
由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*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
[前后消项]
=[n(n+1)(n+2)]/3
所以1^2+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
绛旓細鑷劧鏁骞虫柟鍜屽叕寮:1²+2²+3²+鈥︹+n²=n(n+1)(2n+1)/6 鍘熷紡=100*(100+1)(200+1)/6=338350
绛旓細Sn=1/6*n锛坣+1)(2n+1)绠鍗曟帹瀵艰閾炬帴 鈭村師寮=1/6*100*101*20=338350
绛旓細+ 1 鈥︹2^3 - 1^3 = 3*1^2 + 3*1 + 1 鐩稿姞鍚庯細(n+1)^3 - 1^3 = 3锛1^2 + 鈥︹ + n^2锛+ 3锛1+2+ 鈥︹ + n锛+锛1+鈥︹ +1锛夛細(n+1)^3 - 1^3 = 3锛1^2 + 鈥︹ + n^2锛+ 3锛坣*(n+1)/2锛+n,鏁寸悊鍚庢棦寰.濡傛灉甯埌鎮ㄧ殑璇,(鍙充笂瑙掗噰绾)
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绛旓細涓鐨勫钩鏂瑰姞浜岀殑骞虫柟鍔涓夌殑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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绛旓細n=1鏃讹紝宸﹂潰=1锛屽彸闈=1锛屾垚绔 鍋囪n=k鏃舵垚绔嬶紝鍗1^2+2^2+鈥︹+k^2=[k(k+1锛夛紙2k+1)]/6 鍒欏綋n=k+1鏃讹紝1^2+2^2鈥︹+(k+1)^2 =[k(k+1锛夛紙2k+1)]/6+(k+1)^2 =(k+1){[k(2k+1)+6(k+1)]/6} =(k+1)(2k^2+7k+6)/6 =(k+1)(k+2)(2k+3)/6 =...
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