一的平方一直加到N的平方等于?要过程撒
1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6 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 所以: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 上(n-1)式相加 :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 1的立方加到N的立方的和的公式[n*(n+1)]^2/4绛旓細1鐨勫钩鏂瑰姞鍒皀鐨勫钩鏂鐨勬帹瀵煎叕寮忓涓嬶細1²+2²+3²+鈥︹+n²=n(n+1)(2n+1)/6銆傛牴鎹珛鏂瑰樊鍏紡(a+1)³-a³=3a²+3a+1鍙緱锛宎=1鏃讹細2³-1³=3脳bai1²+3脳1+1锛宎=n鏃讹細(n+1)³-n³=3脳n²+3脳n+1...
绛旓細绛夊紡涓よ竟鐩稿姞锛氾紙n+1)³-1=3锛1²+2²+3²+銆傘傘+n²锛+3锛1+2+3+銆傘傘+n锛+锛1+1+1+銆傘傘+1锛3锛1²+2²+3²+銆傘傘+n²锛=锛坣+1锛³-1-3锛1+2+3+銆傘傘+n锛-锛1+1+1+銆傘傘+1锛3锛1²+2...
绛旓細梅2锛(n+1)=n^3+3n^2+3n+1锛3n^2/2锛3n/2锛峮锛1 =n^3+3/2n^2+n/2 鎵浠,1^2+2^2+3^2+鈥︹+n^2=1/3(n^3+3n^2/2+n/2)=n(n+1)(2n+1)/6 杩欎釜鍏紡鐨勭敤閫斿緢澶,闄や簡鐢ㄤ簬璁$畻杩炵画鑷劧鏁鐨勫钩鏂鍜屽,鍦ㄥ垵楂樹腑鐨勪唬鏁版亽绛夊彉褰腑鏈夌潃寰堝ぇ鐨勪綔鐢.濡傛灉鏄殑璇濆笇鏈涗笉鍚濋噰绾!
绛旓細鍚勭瓑寮忓叏鐩稿姞 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+...
绛旓細杩欎釜鏈夎憲鍚嶇殑 鎺ㄥ鍏紡 Sn=n(n+1)(2n+1)/6 鎺ㄥ鐨勮繃绋嬭鐢ㄦ暟瀛﹀綊绾虫硶 杩欎釜鍏紡璁颁綇鍗冲彲 瑕佽瘉鏄庡拰鍙戠幇鐨勮瘽 鏄釜寰堢箒鐞愮殑杩囩▼ 濡傛灉鏈夊叴瓒g殑璇 浣犲彲浠ョ湅杩欎釜鎺ㄥ杩囩▼ http://wenku.baidu.com/link?url=9XqMICKdNpj3Tg7DwBW34rdeuS202AwZBvvJQikA6qJIbEAEozN6WTD_srdMqEIXOX60ByKWAr_vbWRErV...
绛旓細绛夊紡涓よ竟鐩稿姞锛氾紙n+1)³-1=3锛1²+2²+3²+路路路+n²锛+3锛1+2+3+路路路+n锛+锛1+1+1+路路路+1锛3锛1²+2²+3²+路路路+n²锛=锛坣+1锛³-1-3锛1+2+3+.+n锛-锛1+1+1+.+1锛3锛1²+2²+3...
绛旓細鍏充簬1骞虫柟鍔2骞虫柟鍔3骞虫柟涓鐩村姞鍒皀骞虫柟绛変簬澶氬皯鎬庝箞绠楄繖涓緢澶氫汉杩樹笉鐭ラ亾锛屼粖澶╂潵涓哄ぇ瀹惰В绛斾互涓婄殑闂锛岀幇鍦ㄨ鎴戜滑涓璧锋潵鐪嬬湅鍚э紒1銆1²+2²+3²+鈥︹+n²=n(n+1)(2n+1)/6銆2銆佸彲浠ョ敤(n+1)³-n³=3n²+3n+1绱姞寰楀埌銆3銆佽瘉鏄庤繃绋嬶細鏍规嵁绔嬫柟宸...
绛旓細濂楃敤鍏紡锛1^2+2^2+3^2+4^2+5^2鈥︹︹+n^2=n(n+1)(2n+1)/6銆1^2+2^2+3^2+4^2+5^2鈥︹︹+100^2=100脳101脳201梅6=338 350銆
绛旓細骞虫柟鍜屽叕寮弉(n+1)(2n+1)/6 鍗1^2+2^2+3^2+鈥+n^2=n(n+1)(2n+1)/6 (娉細N^2=N鐨勫钩鏂) 璇佹槑1锛4锛9锛嬧︼紜n^2锛漀(N+1)(2N+1)/6 璇佹硶涓锛堝綊绾崇寽鎯虫硶锛夛細 1銆丯锛1鏃讹紝1锛1锛1锛1锛夛紙2脳1锛1锛/6锛1 2銆丯锛2鏃讹紝1锛4锛2锛2锛1锛夛紙2脳2锛1锛/6锛5 3...
绛旓細1^2+2^2+3^2+鈥︹+n^2=n(n+1)(2n+1)/6 鍒╃敤绔嬫柟宸叕寮 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 ... ...