求证:对任何自然数n,1*2*3...*k+2*3*4...(k+1)+...n(n+1)...(n+k-1)=[n(n+1)...(n+k)]/(k+1)

求证:1*2*3*...*k+2*3*4*...*(k+1)+...+n(n+1)*…*(n+k-1)=[n(n+1)*...*(n+k)]/(k+1) (n为自然数)证一：数学归纳法.略.证二：裂项法.1*2*3...*k = (-0*1*2*3...*k+1*2*3...*k*(k+1))/(k+1)2*3...*k*(k+1)= (-1*2*3.....