求证:2^n-nC1*2^(n-1)+nC2*2^(n-2)+...+nC(n-1)*2+(-1)^n=1
组合恒等式证明n为偶数时nC0+nC2+nC4+……+nCn=nC1+nC3+nC5+……+nCn-1=2^(n-1)我
求证c(n,1)+2c(n,2)+3c(n,3)+...+nc(n,n)=n2^(n-1)
证明:1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+)
计算:C(1,n)+2C(2,n)+3C(3,n) + … + nC(n,n)
求Sn=C(n,1)+2C(n,2)+...+nC(n,n)
2^n/n*(n+1)
求满足不等式C(n,1)+2C(n,2)+……+nC(n,n)
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
(1+2)^n = C(n,0) +2C(n,1) +2^2C(n,2) +2^3C(n,3)+……+2^nC(n,n)
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)+...+
求证:C(0,n)+2C(1,n)+.+(n+1)C(n,n)=2^n+2^(n-1)
设n∈N,n>1.求证:logn (n+1)>log(n+1) (n+2)