一道组合证明题，谢谢！！

设A=C(n,0)+2C(n,1)+……+(n+1)C(n,n)
A=(n+1)C(n,n)+nC(n,n-1)+......+C(n,0)
利用C(n,m)=C(n,n-m)
2A=(n+2)(C(n,0)+C(n,1)+……+C(n,n))=(n+2)*2^n
所以A=2^n+n*2^(n-1)

2[C(n,0)+2C(n,1)+……+(n+1)C(n,n)]=(n+2)(1+1)^n=n2^n+2^(n+1)
所以C(n,0)+2C(n,1)+……+(n+1)C(n,n)
=[n2^n+2^(n+1)]/2=2^n+n*2^(n-1)

太简单了吧这题。
要用到这个两个
C(n,i)=C(n,n-i),i=0.1.2......
C(n,0)+C(n,1)+……+C(n,n)=2^n

C(n,0)+2C(n,1) +……+(n+1)C(n,n) ①
(n+1)C(n,n)+nC(n,n-1)+....+ C(n,0) ②
加起来就是(n+2)[C(n,0)+C(n,1)+……+C(n,n)]=(n+2)2^n
1式与2式是相等的，C(n,0)+2C(n,1)+……+(n+1)C(n,n)＝2^n+n*2^(n-1)