1平方加2平方加3平方一直加到n平方等于多少
1^2+2^2+3^2+..+n^2=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
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
各等式全相加
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
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1锛4锛9锛?锛媥2锛嬶紙x锛1锛2锛漻(x+1)(2x+1)/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銆佺患涓婃墍杩帮紝骞虫柟鍜屽叕寮1^2+2^2...
绛旓細1^2+2^2+3^2+...+n^2=1/6n(n+1)(2n+1)
绛旓細杩欎釜鍏紡瑕佽浣忥細1 + 2 + 3 + ...锛媙 = n(n+1)(2n+1)/6 鎵浠1 + 2 + 3 + ...锛100=锛100脳101脳201锛/6=338350 閲囩撼鍝
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s=1^2+2^2+...+100^2 =100[100+1][2*100+1]/6 =338350 S锛坣)=n(n+1)(2n+1)/6
绛旓細=1*2+2*3+...+n(n+1)-(1+2+...+n)鐢变簬n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3 鎵浠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 [鍓嶅悗娑堥」]=[n(n+1)(n+2)]/3 鎵浠1^2+2^2+3^2+...+n^2 =...
绛旓細鍒╃敤鍏紡锛1^2+2^2+3^3+...+n^2=[n(n+1)(2n+1)]/6 1+2+3+...+n=[n(n+1)]/2 灏卞彲浠ユ眰鍑烘潵浜嗐備竴鐨骞虫柟鍔犱簩鐨骞虫柟鍔犱笁鐨勫钩鏂逛竴鐩村姞鍒浜岄浂1涓冪殑骞虫柟闄や互涓鍔浜屽姞涓変竴鐩村姞鍒2017 =[(2017x2018x4035)/6]/[2017x2018)/2]=4035x2/6 =1345銆
绛旓細3^3 - 2^3 =3*2^2 + 3*2 +1 4^3 - 3^3 =3*3^2 + 3*3 +1 ...(n+1)^3 - n^3=3*n^2 + 3*n +1 浠ヤ笂鍚勫紡鐩稿姞寰楋細(n+1)^3 - 1 = 3(1^2+2^2+3^2+...+n^2) +3(1+2+3+...+n)+n 鎵浠 1^2+2^2+3^2+...+n^2 =1/6*(2n+1)(n+1)n ...
绛旓細姹傝瘉鐨勬柟娉曟湁寰堝锛屾垜浠ュ墠鏄氳繃缁勫悎鏁扮殑瑙勫緥鏉ユ濊冪殑锛2n+1锛夛紙n+1锛塶/6 鎴戜滑鍙互閫氳繃缁勫悎鏁扮殑涓鍏卞叕寮忔潵鑰冭檻锛氾紙n,k锛+(n,k+1)=(n+1,k+1),杩欓噷鐢ㄥ埌鐨勬槸k=2鐨勬儏鍐碉紝鍗筹紙n,2锛+(n,3)=(n+1,3)绠椾竴涓嬪氨鐭ラ亾杩欏叕寮忔槸鍚︽纭簡銆備笅闈㈠彲浠ヨ绠椾簡銆俷^2=2(n+1,2)-n,鎵浠1^2+...
