1的平方加2的平方，加3的平方，一直加到100的平方等于多少？ 一的平方加二的平方加三的平方一直加到一百的平方等于多少
\u4ece1\u7684\u5e73\u65b9\u4e00\u76f4\u52a0\u5230100\u7684\u5e73\u65b9\u662f\u591a\u5c111^2=1*2-1
2^2=2*3-2
.....
.....
n^2=n(n+1)-n
\u7531\u4e8en(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
\u6240\u4ee51*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
[\u524d\u540e\u6d88\u9879]
=[n(n+1)(n+2)]/3
\u6240\u4ee51^2+2^2+3^2+......+n^2
=[n(n+1)(n+2)]/3-[n(n+1)]/2
=n(n+1)[(n+2)/3-1/2]
\u6216\u8005\u6570\u5b66\u5f52\u7eb3\u6cd5..\u6216\u8005
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
\u7b49\u5f0f\u5168\u76f8\u52a0
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+...+n^2)=n(n+1)(2n+1)/6
\u5f53n=100\u65f6(1^2+2^2+...+100^2)=100(100+1)(2*100+1)/6
=100*101*201/6
=50*101*67
=338350
\u89e3\uff1a\u516c\u5f0f\uff1a
1²+2²+3²+....+n²
=1/6 n(n+1)(2n+1)
\u6240\u4ee5
\u53d6n=100\uff0c\u5f97
\u539f\u5f0f=1/6 \u00d7100\u00d7\uff08100+1\uff09\u00d7\uff082\u00d7100+1\uff09
=338350
1的平方加到100的平方
=100×101×201／6=338350
绛旓細1^2+2^2+3^2=1+4+9=14
绛旓細1²+2²+3²=1+4+9 =14
绛旓細瑙ｏ細鍏紡锛1²+2²+3²+...+n²=1/6 n(n+1)(2n+1)鎵浠 鍙杗=100锛屽緱 鍘熷紡=1/6 脳100脳锛100+1锛壝楋紙2脳100+1锛=338350
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1^2+2^2+...+n^2=n(n+1)(2n+1)/6
绛旓細1^2+2^2+3^2 =1+4+9 =14
绛旓細14
绛旓細4銆佺患涓婃墍杩,骞虫柟鍜屽叕寮1^2+2^2+3^2+鈥+n^2=n(n+1)(2n+1)/6鎴愮珛,寰楄瘉.璇佹硶浜岋紙鍒╃敤鎭掔瓑寮(n+1)^3=n^3+3n^2+3n+1锛夛細(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.鎶婅繖...
绛旓細鍒╃敤涓婇潰杩欎釜寮忓瓙鏈夛細2^3 - 1^3 = 3*1^2 + 3*1 + 1 3^3 - 2^3 = 3*2^2 + 3*2 + 1 4^3 - 3^3 = 3*3^2 + 3*3 + 1 5^3 - 4^3 = 3*4^2 + 3*4 + 1 鈥︹(n+1)^3 - n^3 = 3*n^2 + 3n + 1 鎶婁笂杩板悇绛夊紡宸﹀彸鍒嗗埆鐩稿姞 寰楀埌锛(n+1)^3 - 1...
绛旓細涓鐨勫钩鏂瑰姞浜岀殑骞虫柟鍔犱笁鐨勫钩鏂路路路涓鐩村姞鍒皀鐨勫钩鏂 =n(n+1)(2n+1)/6
