1平方加2的平方加3的平方怎么算 1平方加2平方加3平方一直加到n平方等于多少
1\u5e73\u65b9+2\u5e73\u65b9+3\u5e73\u65b9+...+n\u5e73\u65b9\u600e\u4e48\u7b971+\uff082+2\uff09+\uff083+3+3\uff09+...+\uff08n+n+...+n\uff09
n+\uff08n+n-1\uff09+\uff08n+n-1+n-2\uff09+...+\uff08n+n-1+n-2+n-3+...+2+1\uff09
n+\uff08n+n-1\uff09+\uff08n+n-1+n-2\uff09+...+\uff08n+n-1+n-2+n-3+...+2+1\uff09
\u4e09\u4e2a\u76f8\u52a0\u7b49\u4e8e
2n+1+\uff084n+2\uff09+\uff086n+3\uff09+....+n\uff082n+1\uff09
=\uff082n+1\uff09\uff081+2+3+...+n\uff09
=\uff082n+1\uff09\uff081+n\uff09n\u00b7\uff081/2\uff09
\u56e0\u4e3a\u662f\u4e09\u4e2a\u5f0f\u5b50\u76f8\u52a0\u6700\u540e\u8fd8\u8981\u4e58\u4ee51/3\u624d\u662f\u7b54\u6848
=\uff082n+1\uff09\uff081+n\uff09n\u00b7\uff081/2\uff09\u00b7\uff081/3\uff09
\u6269\u5c55\u8d44\u6599\u76f8\u5173\u516c\u5f0f\uff1a
\uff081\uff09\uff08a-b\uff09³\uff1da³\uff0d3a²b\uff0b3ab²-b³
\uff08a-b\uff09³\uff1da³\uff0d3a²b\uff0b3ab²-b³\u7684\u63a8\u5bfc\u8fc7\u7a0b\u5982\u4e0b\uff1a
(a-b)³
=(a-b)(a-b)²\uff08\u5206\u89e3\u6210\u4e24\u4e2a\u56e0\u5f0f\u76f8\u4e58\uff09
=(a-b)(a²-2ab+b²)\uff08\u628a(a-b)²\u7528\u4e58\u6cd5\u8868\u8fbe\u51fa\u6765\uff09
=a³-3a²b+3ab²-b³\uff08\u4f9d\u6b21\u76f8\u4e58\u5f97\u5230\u6700\u540e\u7ed3\u679c\uff09
\uff082\uff09\uff08a+b\uff09³\uff1da³\uff0b3a²b\uff0b3ab²\uff0bb³
\uff083\uff09a³+b³=a³+a²b-a²b+b³=a²\uff08a+b\uff09-b\uff08a²-b²\uff09=a²\uff08a+b\uff09-b\uff08a+b\uff09\uff08a-b\uff09
=\uff08a+b\uff09[a²-b\uff08a-b\uff09]=\uff08a+b\uff09\uff08a²-ab+b²\uff09
\uff084\uff09a³-b³=a³-a²b+a²b-b³=a²\uff08a-b\uff09+b\uff08a²-b²\uff09=a²\uff08a-b\uff09+b\uff08a+b\uff09\uff08a-b\uff09
=\uff08a-b\uff09[a²+b\uff08a+b\uff09]=\uff08a-b\uff09\uff08a²+ab+b²\uff09
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
即1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6 (注：N^2=N的平方)
证明1＋4＋9＋…＋n^2＝N(N+1)(2N+1)/6
证法一（归纳猜想法）：
1、N＝1时,1＝1（1＋1）（2×1＋1）/6＝1
2、N＝2时,1＋4＝2（2＋1）（2×2＋1）/6＝5
3、设N＝x时,公式成立,即1＋4＋9＋…＋x2＝x(x+1)(2x+1)/6
则当N＝x＋1时,
1＋4＋9＋…＋x2＋（x＋1）2＝x(x+1)(2x+1)/6＋（x＋1）2
＝（x＋1）[2（x2）＋x＋6（x＋1）]/6
＝（x＋1）[2（x2）＋7x＋6]/6
＝（x＋1）（2x＋3）（x＋2）/6
＝（x＋1）[（x＋1）＋1][2（x＋1）+1]/6
也满足公式
4、综上所述,平方和公式1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6成立,得证.
证法二（利用恒等式(n+1)^3=n^3+3n^2+3n+1）：
(n+1)^3-n^3=3n^2+3n+1,
n^3-(n-1)^3=3(n-1)^2+3(n-1)+1
.
3^3-2^3=3*(2^2)+3*2+1
2^3-1^3=3*(1^2)+3*1+1.
把这n个等式两端分别相加,得：
(n+1)^3-1=3(1^2+2^2+3^2+.+n^2)+3(1+2+3+...+n)+n,
由于1+2+3+...+n=(n+1)n/2,
代人上式得：
n^3+3n^2+3n=3(1^2+2^2+3^2+.+n^2)+3(n+1)n/2+n
整理后得：
1^2+2^2+3^2+.+n^2=n(n+1)(2n+1)/6
1^2+2^2+3^2
=1+4+9
=14
绛旓細鍏紡锛1²+2²+3²+.+N²=n锛坣+1锛(2n+1)/6 1鐨勫钩鏂瑰姞鍒100鐨勫钩鏂 =100脳101脳201锛6=338350
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
绛旓細脳n梅2-n 6锛1²+2²+3²+路路路+n²锛=2锛坣+1)³-3n锛1+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)鎵浠1²+2²+路路路+n²=n锛坣+1)锛2n+1锛/6銆
绛旓細3* 3^2= 4^3 - 4^2 ...19* 20^2= 20^3 - 20^2 鍘熷紡=2^3 - 2^2+3^3 - 3^2+ 4^3 - 4^2銆傘傘+20^3 - 20^2 =锛1+2^3 +3^3 + 4^3 銆傘傘+20^3 锛-锛1+2^2 +3^2 + 4^2 銆傘傘+20^2锛=銆愶紙20*21锛/2銆慯2-20*锛20+1锛*锛40+1锛/6 =21...
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
绛旓細涓鐨骞虫柟鍔犱簩鐨勫钩鏂瑰姞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
绛旓細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*3^2+2^2-3 4^3-3^3=2*4^2+3^2-4 ...
绛旓細瑕佸垪鍑1鐨骞虫柟鍔2鐨勫钩鏂瑰姞3鐨勫钩鏂涓鐩村姞鍒10鐨勫钩鏂圭殑椤癸紝鎴戜滑鍙互浣跨敤鏁板绗﹀彿鏉ヨ〃绀恒傝繖涓簭鍒楀彲浠ヨ〃绀轰负锛1² + 2² + 3² + 4² + 5² + 6² + 7² + 8² + 9² + 10²绠鍖栦竴涓嬶紝鎴戜滑鍙互鍐欐垚锛氣垜(n²), n浠1鍒10...
绛旓細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 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^...
绛旓細鈥︹︹4^3-3^3=3*3^2+3*3+1 3^3-2^3=3*2^2+3*2+1 2^3-1^3=3*1^2+3*1+1 灏嗕笂闈涓瓑寮忕浉鍔狅紝寰 (n+1)^2-1=3*(1^2+2^2+3^2+鈥︹+n^2)+3(1+2+3+鈥︹+n)+n 鍏朵腑1+2+3+鈥︹+n=n(n+1)/2 甯﹀叆鍓嶅紡鍖栫畝寰 1^2+2^2+3^2+鈥︹+n^2=n(n+1...
