1的平方一直加到N的平方,怎么化简,用什么方法,数学归纳法? 数学，一平方加二平方一直加到n平方，请问如何推出规律？
1\u7684\u5e73\u65b9\u52a0\u5230n\u7684\u5e73\u65b9\u600e\u4e48\u7b97\uff0c\u7528\u6570\u5217\u7684\u65b9\u6cd5\u8fd9\u4e2a\u6709\u8457\u540d\u7684 \u63a8\u5bfc\u516c\u5f0f
Sn=n(n+1)(2n+1)/6
\u63a8\u5bfc\u7684\u8fc7\u7a0b\u8981\u7528\u6570\u5b66\u5f52\u7eb3\u6cd5
\u8fd9\u4e2a\u516c\u5f0f\u8bb0\u4f4f\u5373\u53ef \u8981\u8bc1\u660e\u548c\u53d1\u73b0\u7684\u8bdd \u662f\u4e2a\u5f88\u7e41\u7410\u7684\u8fc7\u7a0b
\u5982\u679c\u6709\u5174\u8da3\u7684\u8bdd \u4f60\u53ef\u4ee5\u770b\u8fd9\u4e2a\u63a8\u5bfc\u8fc7\u7a0b
http://wenku.baidu.com/link?url=9XqMICKdNpj3Tg7DwBW34rdeuS202AwZBvvJQikA6qJIbEAEozN6WTD_srdMqEIXOX60ByKWAr_vbWRErV1EeIGdkxKkdBivLNz3KG1yGu3
Sn=1²+2²+....+n²\uff0c \u662f\u7528\u7acb\u65b9\u6765\u6c42\u548c\u7684\u3002
\u8bb0Tn=1+2+...+n=n(n+1)/2
\u7531\u7acb\u65b9\u5dee\u516c\u5f0f\uff1a(n+1)³-n³=3n²+3n+1
\u4ee3\u5165n=1, 2, ...,n\u5f97\uff1a
2³-1³=3*1²+3*1+1
3³-2³=3*2²+3*2+1
...
(n+1)³-n³=3n²+3n+1
\u4ee5\u4e0an\u4e2a\u5f0f\u5b50\u76f8\u52a0\u5f97\uff1a
(n+1)³-1=3Sn+3Tn+n
\u5316\u7b80\u5373\u5f97\uff1aSn=n(n+1)(2n+1)/6
\u6269\u5c55\u8d44\u6599
\u5e38\u89c1\u6570\u5217\u6c42\u548c\u7684\u65b9\u6cd5\uff1a
1\u3001\u516c\u5f0f\u6cd5\uff1a
\u7b49\u5dee\u6570\u5217\u6c42\u548c\u516c\u5f0f:
Sn=n(a1+an)/2=na1+n(n-1)d/2
\u7b49\u6bd4\u6570\u5217\u6c42\u548c\u516c\u5f0f\uff1a
Sn=na1(q=1)Sn=a1(1-q^n)/(1-q)=(a1-an\u00d7q)/(1-q) (q\u22601)
2\u3001\u9519\u4f4d\u76f8\u51cf\u6cd5
\u9002\u7528\u9898\u578b\uff1a\u9002\u7528\u4e8e\u901a\u9879\u516c\u5f0f\u4e3a\u7b49\u5dee\u7684\u4e00\u6b21\u51fd\u6570\u4e58\u4ee5\u7b49\u6bd4\u7684\u6570\u5217\u5f62\u5f0f { an }\u3001{ bn }\u5206\u522b\u662f\u7b49\u5dee\u6570\u5217\u548c\u7b49\u6bd4\u6570\u5217.
Sn=a1b1+a2b2+a3b3+...+anbn
\u4f8b\u5982\uff1aan=a1+(n-1)d bn=a1\u00b7q^(n-1) Cn=anbn Tn=a1b1+a2b2+a3b3+a4b4.+anbn
qTn= a1b2+a2b3+a3b4+...+a(n-1)bn+anb(n+1)
Tn-qTn= a1b1+b2(a2-a1)+b3(a3-a2)+...bn[an-a(n-1)]-anb(n+1)
Tn(1-q)=a1b1-anb(n+1)+d(b2+b3+b4+...bn) =a1b1-an\u00b7b1\u00b7q^n+d\u00b7b2[1-q^(n-1)]/(1-q) Tn=\u4e0a\u8ff0\u5f0f\u5b50/(1-q)
3\u3001\u88c2\u9879\u6cd5
\u9002\u7528\u4e8e\u5206\u5f0f\u5f62\u5f0f\u7684\u901a\u9879\u516c\u5f0f,\u628a\u4e00\u9879\u62c6\u6210\u4e24\u4e2a\u6216\u591a\u4e2a\u7684\u5dee\u7684\u5f62\u5f0f,\u5373an=f(n+1)\uff0df(n),\u7136\u540e\u7d2f\u52a0\u65f6\u62b5\u6d88\u4e2d\u95f4\u7684\u8bb8\u591a\u9879\u3002
2³-1³=3×1²+3×1+1，
3³-2³=3×2²+3×2+1，
···
(n+1)³-n³=3n²+3n+1.
把上述所有等式左右分别相加，就能得出结果！
原式=1×2+2×3+3×4+....+n×(n+1)-1-2-3-....-n
=1/3×[1×2×3+2×3×4-1×2×3+3×4×5-2×3×4+.....+n×(n+1)×(n+2)-(n-1)×n×(n+1)]-(1+2+3+....+n) )=1/3×n×(n+1)×(n+2)-1/2×n×(n+1)再通分
∵(k+1)³-k³=3k²+3k+1
∴(n+1)³-1³=(3n²+3n+1)+…+(3×1²+3×1+1)=3(n²+…+1²)+3(n+…+1)+n=3(n²+…+1²)+3n(n+1)/2+n
即3(n²+…+1²)=(n+1)³-3n(n+1)/2-(n+1)
∴1²+2²+……+n²=n(n+1)(2n+1)/6
http://zhidao.baidu.com/question/83128584.html?an=0&si=7
=[1*2*3+2*3*4-1*2*3+3*4*5-2*3*4+~~~~~~~~~~~~~+n*(n+1)*(n+2)-(n-1)*n*(n+1)]除以6分之1
绛旓細/6锛嬶紙x锛1锛2 锛濓紙x锛1锛塠2锛坸2锛夛紜x锛6锛坸锛1锛塢/6 锛濓紙x锛1锛塠2锛坸2锛夛紜7x锛6]/6 锛濓紙x锛1锛夛紙2x锛3锛夛紙x锛2锛/6 锛濓紙x锛1锛塠锛坸锛1锛夛紜1][2锛坸锛1锛+1]/6 涔熸弧瓒冲叕寮 4銆佺患涓婃墍杩锛屽钩鏂鍜屽叕寮1^2+2^2+3^2+?+n^2=n(n+1)(2n+1)/6鎴愮珛锛屽緱璇併
绛旓細褰撻渶瑕佽绠椾粠1鐨勫钩鏂瑰埌n鐨勫钩鏂鐨勫拰鏃讹紝鏈変竴涓畝娲佺殑鍏紡锛1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6銆傝繖涓粨璁烘槸閫氳繃鏁板褰掔撼娉曟帹瀵煎緱鍑虹殑锛1.褰搉=1鏃讹紝宸﹁竟鐨勫拰鏄1锛屽彸杈圭殑璁＄畻缁撴灉涔熸槸1锛岄獙璇佷簡鍏紡鍦╪=1鏃舵垚绔嬨2.鎺ヤ笅鏉ュ亣璁緉=x鏃讹紝鍏紡姝ｇ‘锛屽嵆1+4+9+...+x^2=x(x+1)(...
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鏃锛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銆佽...
绛旓細n=1 1鐨勫钩鏂=1,锛1+1锛*1*锛2+1锛/6=1 鎵浠ュ綋n=k (k+1)*k*(2k+1)/6=1鏂+2鏂+.+k鏂 n=k+1涔熸垚绔 1f+2f+3f+...+kf+(k+1)f =(k+1)*k*(2k+1)/6+(k+1)f =(k+1)*k*(2k+1)/6+6(k+1)f/6 =(k+1+1)*(k+1)*[2(k+1)+1]/6 鐢涓鍙煡,鍛介...
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绛旓細涓鐨勫钩鏂瑰姞浜岀殑骞虫柟鍔犱笁鐨勫钩鏂孤仿仿蜂竴鐩村姞鍒皀鐨勫钩鏂 =n(n+1)(2n+1)/6
绛旓細/6锛嬶紙x锛1锛2 锛濓紙x锛1锛塠2锛坸2锛夛紜x锛6锛坸锛1锛塢/6 锛濓紙x锛1锛塠2锛坸2锛夛紜7x锛6]/6 锛濓紙x锛1锛夛紙2x锛3锛夛紙x锛2锛/6 锛濓紙x锛1锛塠锛坸锛1锛夛紜1][2锛坸锛1锛+1]/6 涔熸弧瓒冲叕寮 4銆佺患涓婃墍杩锛屽钩鏂鍜屽叕寮1^2+2^2+3^2+?+n^2=n(n+1)(2n+1)/6鎴愮珛锛屽緱璇併
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n^3-(n-1)^3=2*n^2+(n-1)^2-n 鍚勭瓑寮忓叏鐩稿姞 n^3-1^3=2*(2^2+3^2+.+n^2)=n^3+n^2+n(n+1)/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^...
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
