C(0,0)+1/2C(1,n)+1/3C(2,n)+…1/kC(k-1,n)…+1/(n+1)C(n,n)=1/(n+1)
问题描述：
C(0,0)+1/2C(1,n)+1/3C(2,n)+…1/kC(k-1,n)…+1/(n+1)C(n,n)=1/(n+1)
答
证明：【1/i】C(i-1,n)=1/i*n!/(i-1)!*(n+1-i)!)=n!/(i!*(n+1-i)!)
=【1/(n+1)】(n+1)!/(i!*(n+1-i)!)=1/(n+1)C(i,n+1)
C(0,0)+1/2C(1,n)+1/3C(2,n)+…1/kC(k-1,n)…+1/(n+1)C(n,n)
=1/(n+1)(C(0,n+1)+C(1,n+1)....+C(n,n+1))
=(2^(n+1)-1)/(n+1)
稍有改动——应该更清楚了。
答
证明：1/iC(i-1,n)=1/i*n!/(i-1)!*(n+1-i)!)=n!/(i!*(n-1+i)!)=1/(n+1)(n+1)!/(i!*(n+1-i)!)=1/(n+1)C(i,n+1)C(0,0)+1/2C(1,n)+1/3C(2,n)+…1/kC(k-1,n)…+1/(n+1)C(n,n)=1/(n+1)(C(0,n+1)+C(1,n+1).+C(n,n+1))=(2^(...