组合数公式的题c(n,1)+2c(n,2)+...+nc(n,n) = n[c(n-1,0)+c(n-1,1)+...+c(n-1_数学解答

组合数公式的题
c(n,1)+2c(n,2)+...+nc(n,n) = n[c(n-1,0)+c(n-1,1)+...+c(n-1,n-1)]=n2^n-1

人气：490 ℃　时间：2020-04-13 04:14:22

解答

设Sn=c(n,1)+2c(n,2)+...+nc(n,n)---(1)
c(n,m)=c(n,n-m)
倒写Sn得
Sn=nc(n,n)+(n-1)c(n,n-1)+...2c(n,2)+c(n,1)---(2)
(1)+(2)得
2Sn=n(c(n,0)+c(n,1)+...c(n,n-1)+c(n,n))=n*2^n
Sn=n*2^n-1