1^2+2^2...+n^2=n(n+1)(2n+1)/6 这条公式怎么来的?

来源：学生作业帮助网编辑：作业帮时间：2024/07/11 02:14:02

x��j�@�_��n6ɒ��)��}1�>��r��^�s�@��HM�[tf7*Qz��|��'lVI J��X�r+�J��wܽ7��i{�|k�ׯ��f��φ��D߭�U��K�B�}��n�^&R �:C(O� PY�<݊`$ ��\O�~�Z~|�@�Jϴq\RB�� BR�3��ͧ�$v¼<7)�Tl��>vBL��$x�9��r�~�}�LO�[��p��

1^2+2^2...+n^2=n(n+1)(2n+1)/6 这条公式怎么来的?
1^2+2^2...+n^2=n(n+1)(2n+1)/6 这条公式怎么来的?

1^2+2^2...+n^2=n(n+1)(2n+1)/6 这条公式怎么来的?
1^2+2^2+3^2+……+n^2
=(1^2+1)+(2^2+2)+(3^2+3)+……+(n^2+n)-n(n+1)/2
=2[(2*1)/2+(3*2)/2+(4*3)/2+……+n*(n+1)/2]-n(n+1)/2
=2(C22+C32+C42+……+C(n+1)2)-n(n+1)/2,(C22表式C2选2,C32表式C3选2……)
=2(C33+C32+C42+……+C(n+1)2))-n(n+1)/2
=2C(n+2)3)-n(n+1)/2,(C33+C32=C43,C43+C42=C53……)
=(n+1)n(n-1)/3-n(n+1)/2
=[2(n+2)(n+1)n-3n(n+1)]/6
=n(n+1)(2n+1)/6

2^n/n*(n+1) 证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n f(x)=e^x-x 求证(1/n)^n+(2/n)^n+...+(n/n)^n 当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)...1=n (n+2)!/(n+1)! 2n*1/2^n=n/2^(n-1)? 阶乘(2n-1)!=（2n)!/(2^n*n! 为什么n!/(2!(n-2)!)=n(n-1)/2! 高数题：n趋近于0,lim｛1/（n^2+n+1)+2/(n^2+n+2)+3/(n^2+n+3)+.+n/(n^2+n+n)｝=? Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗? 为什么:n×(n+1)=1/3[n(n+1)×(n+2)-(n-1)×n×(n+1)] 为什么n(n+1)(n+2)(n+3)=(n²+3n+2)(n²+3n)? n(n+1)/1+(n+1)(n+2)/1+.+(n+2007)(n+2008)/1=? 数列a(n)=n (n+1)(n+2)(n+3),求S(n) n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10这道题怎么解 n(n-1)(n-2)（n-3)（n-4)=154440.求N值要步骤证明:c(n,0)c(n,1)+c(n,1)c(n,2)+...c(n,n-1)c(n,n)=c(2n,n-1) [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简